Aniline, a starting compound for urethane plastic foams, consists of C, H, and N.
Combustion of such compounds yields CO2 (carbon dioxide), H2O (water), and N2 as
products. If the combustion of 9.71 mg of Aniline yields 6.63 mg of H2O and 1.46 mg of
N2, what is its empirical formula? The molecular weight of Aniline is 93 amu. What is its
molecular formula?

Answers

Answer 1

The empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.

Steps

To determine the empirical formula of Aniline, we need to calculate the number of moles of each element present in the given mass of the compound and then find the smallest whole-number ratio between them.

Given:

Mass of Aniline = 9.71 mg

Mass of water produced = 6.63 mg

Mass of N2 produced = 1.46 mg

The molecular weight of Aniline = 93 amu

First, let's calculate the number of moles of water and nitrogen produced:

moles of H2O = 6.63 mg / 18.015 g/mol = 0.3680 mmol

moles of N2 = 1.46 mg / 28.014 g/mol = 0.0521 mmol

Next, we need to calculate the number of moles of carbon and nitrogen present in Aniline:

moles of C = (9.71 mg - (0.3680 mmol x 12.011 g/mol)) / 12.011 g/mol = 0.4811 mmol

moles of N = 0.0521 mmol

Now, we need to find the smallest whole-number ratio between these elements. We can divide the number of moles of each element by the smallest value, which is 0.0521 mmol:

moles of C = 0.4811 mmol / 0.0521 mmol = 9.231 ≈ 9

moles of N = 1

moles of H = not calculated, but we can find it using the difference in mass between Aniline and the products formed.

Now, let's calculate the mass of carbon and nitrogen in Aniline:

mass of C = 9 x 12.011 g/mol = 108.099 g/mol

mass of N = 1 x 14.007 g/mol = 14.007 g/mol

Finally, we can calculate the mass of hydrogen by taking the difference between the mass of Aniline and the sum of the masses of carbon, nitrogen, water, and nitrogen:

mass of H = 9.71 mg - (108.099 g/mol + 14.007 g/mol + 6.63 mg + 1.46 mg) = 2.47 mg

Now, we can calculate the number of moles of hydrogen:

moles of H = 2.47 mg / 1.008 g/mol = 2.449 mmol

Finally, we can express the empirical formula of Aniline as: C9H10N

To find the molecular formula, we need to calculate the molecular weight of the empirical formula:

Empirical formula weight = (9 x 12.011 g/mol) + (10 x 1.008 g/mol) + (1 x 14.007 g/mol) = 93.126 g/mol

Now, we can find the molecular formula by dividing the molecular weight of Aniline by the empirical formula weight and multiplying each subscript in the empirical formula by the result:

The molecular weight of Aniline = 93 amu

The molecular weight of the empirical formula = 93.126 g/mol

Molecular formula = empirical formula x (Molecular weight of Aniline / Empirical formula weight)

= C9H10N x (93 amu / 93.126 g/mol)

= C9H9.9N

Therefore, the empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.

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Related Questions

20points!!!
I need ASAP!!!
You are given 650.0 g of a 0.75 molal solution of methanal dissolved in water. How many grams of methanal are in this amount of solution?

Answers

mass of methanal = -350.0 g

This result is still negative, which means that the given information cannot be correct.

We are given a 0.75 molal solution of methanal (also known as formaldehyde) dissolved in water, and we are told that the total mass of the solution is 650.0 g. We need to calculate the mass of methanal in this solution.

Assuming this information, we can proceed with the calculation as follows:

The molar mass of methanal (HCHO) is 30.03 g/mol. A 0.75 molal solution of methanal means that 0.75 moles of methanal are dissolved in 1 kg of water.

First, we need to calculate the mass of water in the solution. We can use the molality formula to do this:

molality = moles of solute / mass of solvent (in kg)

0.75 molal solution means 0.75 moles of methanal are dissolved in 1 kg of water.

Therefore, the amount of water in the solution is:

moles of methanal = 0.75 mol

mass of water = moles of methanal / molality = 0.75 mol / 0.75 mol/kg = 1 kg

Since the total mass of the solution is 650.0 g, the mass of methanal can be found by subtracting the mass of water from the total mass:

mass of methanal = total mass of solution - mass of water

mass of methanal = 650.0 g - 1000 g

mass of methanal = -350.0 g

This result is still negative, which means that the given information cannot be correct.

Write the net ionic equation for the following molecular equation. 2FeCl3(aq) + 3(NH4)2CO3(aq) Fe2(CO3)3(s) + 6NH4Cl(aq)

Answers

The first step is to write the complete ionic equation by breaking apart all soluble ionic compounds into their constituent ions:

2Fe³⁺(aq) + 6Cl⁻(aq) + 6NH₄⁺(aq) + 3CO₃²⁻(aq) → Fe₂(CO₃)₃(s) + 6NH₄⁺(aq) + 6Cl⁻(aq)

Next, we cancel out any spectator ions that appear on both sides of the equation, which in this case is the 6NH₄⁺ and the 6Cl⁻.

The resulting net ionic equation is:

2Fe³⁺(aq) + 3CO₃²⁻(aq) → Fe₂(CO₃)₃(s)

Therefore, the net ionic equation for the given molecular equation is:

2Fe³⁺(aq) + 3CO₃²⁻(aq) → Fe₂(CO₃)₃(s)


Without checking and reporting their boiling points, rank the boiling points of H2O, MgCl2, HBr, and HF H2S from the lowest to highest based on your knowledge of intermolecular forces, hydrogen bond, and ionic bond. Explain how you make your predictions without checking a reference.

Answers

As HF is hydrogen bonded, it has a highest boiling point of 19 oC or less, when it becomes liquid. The boiling points of the remaining hydrogen halides, which are gaseous, are determined by the van der Waal forces.

H2O or HF, which has the higher boiling point?

Hydrogen fluoride (H 2 F) has a lower boiling point than water (H 2 O) ( ). H2O has a boiling point of 100 °C while only having a boiling point of 5 °C. Despite the fact that water and hydrogen fluoride have stronger hydrogen bonds than each other, water only forms two hydrogen bonds per molecule whereas hydrogen fluoride only forms one.

Which has the highest boiling point between H2O and HCL?

Furthermore, water creates hydrogen bonds between nearby molecules. The extremely electronegative oxygen on the other molecule interacts with hydrogen. This is a strong type of interaction that takes more energy than dipole-dipole interactions. As a result, water boils at a greater temperature than HCL.

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If you have 7.22 x 1023 atoms of chromium (Cr), how many moles of chromium do you
have?

O.98 moles Cr
O 1.62 moles Cr
O 1.36 moles Cr
O 1.19 moles Cr

Answers

Answer:

The correct answer is 1.19 moles Cr

Explanation:

We can use Avogadro's number, which is 6.022 x 10^23 particles (atoms or molecules) per mole, to convert the given number of atoms to moles.

Number of moles = Number of atoms / Avogadro's number

Number of moles = 7.22 x 10^23 / 6.022 x 10^23

Number of moles = 1.199 moles (rounded to three significant figures)

Therefore, the number of moles of chromium is approximately 1.19 moles Cr.

Answer:

Explanation:

To calculate the number of moles of chromium, we need to use Avogadro's number, which is defined as the number of atoms in 12 grams of carbon-12. Avogadro's number is approximately 6.022 x 10^23 atoms per mole.

Given that we have 7.22 x 10^23 atoms of chromium, we can calculate the number of moles of chromium using the following formula:

moles of chromium = number of atoms of chromium / Avogadro's number

moles of chromium = 7.22 x 10^23 / 6.022 x 10^23

moles of chromium ≈ 1.20

Therefore, we have approximately 1.20 moles of chromium.

What is the formula for a compound composed of Cr³+ ions and SO32- ions?

Answers

Answer:

The compound composed of Cr³+ ions and SO32- ions is called chromium(III) sulfite.

The formula for this compound can be determined by balancing the charges of the Cr³+ ion and the SO32- ion so that the overall charge of the compound is neutral.

The charge of the Cr³+ ion is +3, while the charge of the SO32- ion is -2. To balance these charges, we need two SO32- ions for every Cr³+ ion:

Cr³+ + 2 SO32- → Cr2(SO3)3

Therefore, the formula for the compound is Cr2(SO3)3.

How to solve this problem?

Answers

1.) 4.11 grams of CaC2 will produce 1.67 grams of C2H2 ; 2.) 17.01 grams of Ca(OH)2 will produce 14.74 grams of CaC2.

What is chemical reaction?

Process that leads to the chemical transformation of one set of chemical substances to another is called chemical reaction.

Balanced chemical equation for the reaction is:

CaC2 + 2H2O → C2H2 + Ca(OH)2

Molar mass of CaC2 is:

Ca = 40.08 g/mol

C = 12.01 g/mol

2 x H = 2 x 1.01

Total = 64.1 g/mol

So 4.11 g of CaC2 is equal to: n = m/M = 4.11 g / 64.1 g/mol = 0.064 mol

Molar mass of C2H2 is:

2 x C =2 x 12.01

2 x H =2 x 1.01

Total = 26.04 g/mol

So mass of C2H2 produced is:

m = n x M = 0.064 mol x 26.04 g/mol = 1.67 g

Therefore, 4.11 grams of CaC2 will produce 1.67 grams of C2H2.

Molar mass of Ca(OH)2 is:

Ca = 40.08 g/mol

2 x O=2 x 16.00

2 x H=2 x 1.01

Total = 74.10 g/mol

So 17.01 g of Ca(OH)2 is equal to: n = m/M = 17.01 g / 74.10 g/mol = 0.2297 mol

Molar mass of CaC2 is:

Ca = 40.08 g/mol

2 x C=2x12.01

Total = 64.1 g/mol

So the mass of CaC2 produced is:

m = n x M = 0.2297 mol x 64.1 g/mol = 14.74 g

Therefore, 17.01 grams of Ca(OH)2 will produce 14.74 grams of CaC2.

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PLEASE HELP AND SHOW YOUR WORK

1) how many grams of cl2 are produced from 45 grams of aucl3?

2) calculate the grams of agcl produced from 4.75 grams alcl3.

3) calculate the mass of naoh needed to produce 25.2 grams of na2so4

4) calculate the mass of co2 produced from 25 grams of o2

Answers

1. 15.78 grams of Cl2 are produced from 45 grams of AuCl3.

2. 15.31 grams of AgCl are produced from 4.75 grams of AlCl3.

3. 3.55 grams of NaOH are needed to produce 25.2 grams of Na2SO4.

4.  34.40 grams of CO2 are produced from 25 grams of O2.

Steps

1. To determine how many grams of Cl2 are produced from 45 grams of AuCl3, we need to use the balanced chemical equation for the reaction:

2 AuCl3 + 3 Cl2 → 2 AuCl5

From this equation, we can see that 3 moles of Cl2 are produced for every 2 moles of AuCl3. To calculate the number of moles of AuCl3 in 45 grams, we divide the mass by the molar mass:

45 g AuCl3 x (1 mol AuCl3 / 303.33 g AuCl3) = 0.1482 mol AuCl3

Using the mole ratio from the balanced chemical equation, we can calculate the number of moles of Cl2 produced:

0.1482 mol AuCl3 x (3 mol Cl2 / 2 mol AuCl3) = 0.2223 mol Cl2

Finally, we can convert the number of moles of Cl2 to grams by multiplying by the molar mass:

0.2223 mol Cl2 x 70.906 g/mol = 15.78 g Cl2

Therefore, 15.78 grams of Cl2 are produced from 45 grams of AuCl3.

2. The balanced chemical equation for the reaction between aluminum chloride (AlCl3) and silver nitrate (AgNO3) is:

3 AgNO3 + AlCl3 → 3 AgCl + Al(NO3)3

From this equation, we can see that 3 moles of AgCl are produced for every 1 mole of AlCl3. To calculate the number of moles of AlCl3 in 4.75 grams, we divide the mass by the molar mass:

4.75 g AlCl3 x (1 mol AlCl3 / 133.34 g AlCl3) = 0.0356 mol AlCl3

Using the mole ratio from the balanced chemical equation, we can calculate the number of moles of AgCl produced:

0.0356 mol AlCl3 x (3 mol AgCl / 1 mol AlCl3) = 0.1068 mol AgCl

Finally, we can convert the number of moles of AgCl to grams by multiplying by the molar mass:

0.1068 mol AgCl x 143.32 g/mol = 15.31 g AgCl

Therefore, 15.31 grams of AgCl are produced from 4.75 grams of AlCl3.

3. The balanced chemical equation for the reaction between NaOH and Na2SO4 is:

2 NaOH + Na2SO4 → 2 Na2SO4 + 2 H2O

From the balanced equation, we see that 2 moles of NaOH react with 1 mole of Na2SO4 to produce 2 moles of Na2SO4 and 2 moles of H2O. To calculate the mass of NaOH needed to produce 25.2 grams of Na2SO4, we need to use the molar mass of Na2SO4:

Molar mass of Na2SO4 = 142.04 g/mol

First, we need to calculate the number of moles of Na2SO4 produced from 25.2 grams:

25.2 g Na2SO4 x (1 mol Na2SO4 / 142.04 g Na2SO4) = 0.1774 mol Na2SO4

Since 2 moles of NaOH react with 1 mole of Na2SO4, we need half that amount, or 0.0887 mol NaOH, to react completely with the given amount of Na2SO4. Finally, we can calculate the mass of NaOH needed using its molar mass:

0.0887 mol NaOH x 40.00 g/mol = 3.55 g NaOH

Therefore, 3.55 grams of NaOH are needed to produce 25.2 grams of Na2SO4.

4. To determine the mass of CO2 produced from 25 grams of O2, we need to use the balanced chemical equation for the combustion of carbon:

C + O2 → CO2

From this equation, we can see that 1 mole of O2 reacts with 1 mole of carbon to produce 1 mole of CO2. To calculate the number of moles of O2 in 25 grams, we divide the mass by the molar mass:

25 g O2 x (1 mol O2 / 32.00 g O2) = 0.78125 mol O2

Since the mole ratio of O2 to CO2 is 1:1, we know that 0.78125 moles of CO2 are produced. Finally, we can convert the number of moles of CO2 to grams by multiplying by the molar mass:

0.78125 mol CO2 x 44.01 g/mol = 34.40 g CO2

Therefore, 34.40 grams of CO2 are produced from 25 grams of O2.

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Ammonium phosphate ((NH4)3PO4) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H3PO4) with ammonia (NH3).
What mass of ammonium phosphate is produced by the reaction of 7.73 g of ammonia?

Answers

Answer:

= 22.51 g of ammonium phosphate

Step-by-step explanation:

The balanced chemical equation for the reaction between phosphoric acid and ammonia to form ammonium phosphate is:

H3PO4 + 3 NH3 → (NH4)3PO4

From the equation, we can see that 3 moles of ammonia react with 1 mole of phosphoric acid to produce 1 mole of ammonium phosphate.

To determine the mass of ammonium phosphate produced from 7.73 g of ammonia, we need to first calculate the number of moles of ammonia, and then use the mole ratio from the balanced equation to find the number of moles of ammonium phosphate. Finally, we can convert the number of moles of ammonium phosphate to its mass using its molar mass.

The molar mass of NH3 is 17.03 g/mol, and the molar mass of (NH4)3PO4 is 149.09 g/mol.

Number of moles of NH3 = mass / molar mass = 7.73 g / 17.03 g/mol = 0.453 mol

From the balanced equation, 3 moles of NH3 react to form 1 mole of (NH4)3PO4.

Number of moles of (NH4)3PO4 = 0.453 mol NH3 / 3 mol NH3/mol (NH4)3PO4 = 0.151 mol (NH4)3PO4

Mass of (NH4)3PO4 = number of moles x molar mass = 0.151 mol x 149.09 g/mol = 22.51 g

Therefore, 7.73 g of ammonia reacts to form 22.51 g of ammonium phosphate.

Which of the steps in prokaryotic binary fission is correct?
Question 11 options:

a) All of these choices are correct.
b) The two replicated chromosomes remain attached to the plasma membrane.
c) The cell continues to grow outward symmetrically, separating the two chromosomes.
d) Cell wall material is laid down at the midpoint to separate the two daughter cells.
e) DNA is replicated bidirectionally from a single point on the circular chromosome.

Answers

From a single location on the circular chromosome, DNA is copied in both directions.

The correct option is E.

What is of binary fission?

Binary fission is the process of asexual reproduction in which one organism is divided into two separate ones. An organism's genetic material, or deoxyribonucleic acid (DNA), doubles when it splits into two halves (cytokinesis) through binary fission, with each new species inheriting one copy of the latter.

What cells use binary fission?

Bacterial binary fission is the method by which bacteria split their cells. Find out how binary fission functions, including how to make a new cell wall and a copy of a bacterial chromosome. Mitosis and binary fission, two types of asexual reproduction, both entail the division of a parent cell into two identical daughter cells.

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Chemistry Help Please!

1. Classify the following reactions as a redox or precipitate reaction. If it is a precipitate reaction, then specify whether it is a synthesis, decomposition, single displacement, or combustion reaction.

a. Ca(s) + Br2(L) ⟶ CaBr2(s)
b. C6H12(L) + 9O2 ⟶ 6CO2(g) + 6H2O
c. H2O(g) + C(s) ⟶ CO(g) + H2(g)
d. Pb(NO3)2(aq) + H2SO4(aq) ⟶ PbSO4(s) + 2HNO3(aq)
e. 2KClO3(s) ⟶ 2KCl(s) + 3O2(g)

2. Identify the type of reaction, write the formula for the product, and then balance the reaction.
a. Mg(s) + I2(s) ⟶
b. Ba(s) + HBr(aq) ⟶
c. K(s) + H2O(L) ⟶
d. H3PO4(aq) + CaCl2(aq) ⟶
e. C6H12(L) + O2(g) ⟶

Answers

The solution to the given questions on the various chemical reactions , and the balancing of the equations are given below in the explanation below, for clarity purpose

Types of Chemical Reaction

Redox (reduction-oxidation) reactions involve the transfer of electrons between reactants. In a redox reaction, one reactant undergoes oxidation (loses electrons) while another reactant undergoes reduction (gains electrons). These reactions are important in many biological and industrial processes.

Combustion reactions are exothermic reactions that involve the rapid oxidation of a substance in the presence of oxygen, resulting in the release of energy in the form of heat and light. Common examples of combustion reactions include the burning of fossil fuels such as coal, oil, and natural gas.

A precipitation reaction is a type of double displacement reaction in which two aqueous solutions are mixed, and a solid product (a precipitate) forms. Double displacement reactions involve the exchange of ions between two compounds in solution. Precipitation reactions are often used in chemical analysis to identify the presence of certain ions in a solution.

Decomposition reactions involve the breakdown of a single compound into two or more smaller compounds or elements. These reactions can be initiated by heat, light, or other catalysts. For example, the decomposition of hydrogen peroxide into water and oxygen gas is a common laboratory demonstration of a decomposition reaction.

The solution to the given questions are:

1

a. Redox reaction

b. Combustion reaction

c. Redox reaction

d. Precipitate reaction, double displacement

e. Decomposition reaction

2

a. Synthesis reaction: Mg(s) + I2(s) ⟶ MgI2(s)

b. Single displacement reaction: Ba(s) + 2HBr(aq) ⟶ BaBr2(aq) + H2(g)

c. Single displacement reaction: K(s) + H2O(L) ⟶ KOH(aq) + H2(g)

d. Double displacement reaction: 3H3PO4(aq) + 2CaCl2(aq) ⟶ Ca3(PO4)2(s) + 6HCl(aq)

e. Combustion reaction: C6H12(L) + 9O2(g) ⟶ 6CO2(g) + 6H2O(g)

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Answer:

1. a. This is a redox reaction because Ca is oxidized from an oxidation state of 0 to +2, while Br2 is reduced from an oxidation state of 0 to -1.

b. This is a combustion reaction because a hydrocarbon (C6H12) reacts with oxygen to produce carbon dioxide and water.

c. This is a redox reaction because C is oxidized from an oxidation state of 0 to +2, while H2O is reduced from an oxidation state of +1 to 0.

d. This is a precipitation reaction and a double displacement reaction. PbSO4 is a solid precipitate that forms, while HNO3 and H2SO4 remain in solution.

e. This is a decomposition reaction because KClO3 decomposes into KCl and O2.

2. a. This is a single displacement reaction. The formula for the product is MgI2. The balanced reaction is: Mg(s) + I2(s) ⟶ MgI2(s)

b. This is a single displacement reaction. The formula for the product is BaBr2. The balanced reaction is: Ba(s) + 2HBr(aq) ⟶ BaBr2(aq) + H2(g)

c. This is a synthesis reaction. The formula for the product is KOH. The balanced reaction is: 2K(s) + 2H2O(l) ⟶ 2KOH(aq) + H2(g)

d. This is a double displacement reaction. The formula for the product is H3PO4 + CaCl2 ⟶ Ca3(PO4)2 + 6HCl. The balanced reaction is: 2H3PO4(aq) + 3CaCl2(aq) ⟶ Ca3(PO4)2(s) + 6HCl(aq)

e. This is a combustion reaction. The formula for the product is CO2 + H2O. The balanced reaction is: C6H12(l) + 9O2(g) ⟶ 6CO2(g) + 6H2O(l)

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What is the molarity (M) of a solution made by dissolving 75 g of Epsom salt (MgSO,) in 2.3 Liters of solution?

SHOW WORK ​

Answers

The molarity of the Epsom salt solution is 0.271 M.

Steps

To find the molarity (M) of the Epsom salt (MgSO4) solution, we need to first calculate the number of moles of MgSO4 present in the given mass of Epsom salt. We can use the formula:

moles = mass / molar mass

where the molar mass of MgSO4 is 120.37 g/mol.

mass of MgSO4 = 75 g

molar mass of MgSO4 = 120.37 g/mol

moles of MgSO4 = 75 g / 120.37 g/mol = 0.6237 mol

Next, we need to calculate the volume of the solution in liters, since molarity is defined as the number of moles of solute per liter of solution:

volume of solution = 2.3 L

Now we can calculate the molarity of the Epsom salt solution:

Molarity = moles of MgSO4 / volume of solution

= 0.6237 mol / 2.3 L

= 0.271 M

Therefore, the molarity of the Epsom salt solution is 0.271 M.

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Calculate the amount of heat needed to melt 148. g of solid ethanol and bring it to a temperature of -26.8C. Be sure your answer has a unit symbol and the correct number of significant digits.

Answers

Explanation:

The amount of heat required to melt a solid substance is given by the formula:

Q = m * ΔHf

where Q is the amount of heat required, m is the mass of the substance, and ΔHf is the heat of fusion of the substance. The heat of fusion of ethanol is 111.3 J/g.

So, for melting 148 g of solid ethanol, the amount of heat required is:

Q1 = 148 g * 111.3 J/g = 16499.6 J

The amount of heat required to raise the temperature of the liquid ethanol from -26.8°C to its melting point of -114.1°C is given by the formula:

Q2 = m * C * ΔT

where Q2 is the amount of heat required, m is the mass of the substance, C is its specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of ethanol is 2.44 J/g°C. The change in temperature is:

ΔT = (-114.1°C) - (-26.8°C) = -87.3°C

Since we want to bring the ethanol to its melting point, we assume that it is still in the liquid state. Therefore, we use the specific heat capacity of liquid ethanol for this calculation.

So, for raising the temperature of 148 g of liquid ethanol from -26.8°C to -114.1°C, the amount of heat required is:

Q2 = 148 g * 2.44 J/g°C * (-87.3°C) = -30834.8 J

Note that the negative sign indicates that heat is lost from the ethanol as it cools.

The total amount of heat required to melt 148 g of solid ethanol and bring it to a temperature of -26.8°C is the sum of Q1 and Q2:

Qtotal = Q1 + Q2 = 16499.6 J + (-30834.8 J) = -14335.2 J

The answer is -14335 J (rounded to three significant digits), which means that 14335 J of heat energy must be removed from the system (or the surroundings must supply 14335 J of heat energy) to carry out the process as described.

What mass of magnesium bromate will have the same number of oxygen atoms as 10.00 g of sodium bromite?

Answers

Around 6.14 x [tex]10^{24}[/tex] g of magnesium bromate has the same mass as 10.00 g of sodium bromite and the same amount of oxygen atoms.

How much magnesium bromate needs to be produced to have 10,000 g of sodium bromite's amount of oxygen atoms?

Writing down the chemical formulas for magnesium bromate and sodium bromite is the first step in solving this issue. Next, we must determine the mass of magnesium bromate that has the same amount of oxygen atoms as 10.00 g of sodium bromite using the mole ratio.

The compounds' chemical formulae are as follows:

Bromate of magnesium: Mg (BrO3)2

Bromite of sodium: NaBrO2

The molar mass and Avogadro's formula can be used to determine how many oxygen atoms are present in 10.00 g of sodium bromite.

NaBrO2 has a molar mass of 102.89 g/mol.

10.00 g divided by 102.89 grammes per mole yields 0.097 moles.

The formula for the number of oxygen atoms in 0.097 mol is 0.097 mol x 2 mol O / 1 mol. O = 0.194 mol NaBrO2

In 10.00 g of NaBrO2, there are 1.17 x [tex]10^{23}[/tex] oxygen atoms, which is equal to 0.194 mol x 6.022 x [tex]10^{23}[/tex] molecules/mol.

Mg(BrO3)2 molecules each contain six oxygen atoms.

1.17 x [tex]10^{23}[/tex] O atoms divided by 6 molecules' worth of Mg(BrO3)2 results in 1.95 x [tex]10^{22}[/tex] molecules of Mg(BrO3)2.

1.95 x [tex]10^{22}[/tex] molecules of Mg(BrO3)2 weigh 6.14 x [tex]10^{24}[/tex] g when multiplied by the molecular weight of 314.70.

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Use the activity series to predict which of the following single displacement reactions would take place. For those that occur, write the products and the number of electrons transferred. For those that do not, write NR (to indicate no reaction).

a) Al(s) + SnSO4(aq) →
b) Cu(s) + Zn(SO4) (aq) →
c) Mg(s) + Fe(NO3)2(aq) →

#Electrons
Transferred
a)
b)
c)

Answers

a) Al(s) + SnSO4(aq) → six electrons transferred

b) Mg(s) + Fe(NO3)2(aq) → Two electrons transferred

What is the activity series?

The activity series is a list of metals and nonmetals in order of their relative reactivity. It is a useful tool for predicting the outcome of single displacement reactions and for understanding the behavior of metals and nonmetals in various chemical reactions.

In the activity series, the most reactive metals are listed at the top, while the least reactive metals are listed at the bottom. This means that the metals at the top of the activity series are more likely to lose electrons and undergo oxidation, while the metals at the bottom are less likely to lose electrons and undergo oxidation. Similarly, nonmetals are listed in order of their relative ability to gain electrons and undergo reduction.

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The perimeter of the opposite figure:​

Answers

Perimeter is the total length of the boundary of a closed figure. The perimeter of a simple ... We know that the opposite sides of a rectangle are equal.

[tex] \: [/tex]

Peroxyacyl nitrates contain two relatively weak bonds that can be broken by low-energy photons. The weakest is the N−O
bond in the nitrate functional group of the molecule with a standard bond-dissociation energy of 201 kJ mol−1
. What is the longest wavelength of a photon, in nanometres, that has enough energy to cause the photodissociation of the weakest bond. Assume all energy of the photon is efficiently used for the photodissociation.

Answers

The longest wavelength of a photon that has enough energy to cause the photodissociation of the weakest bond in a peroxyacyl nitrate molecule is approximately 590 nm.

Given the weakest bond in Peroxyacyl nitrate is N-O bond.

The standard bond-dissociation energy of N-O = 201 kJ mol−1

Let the wavelength = l

We know that E = h c / λ, where E is the energy in joules, h is Planck's constant, c is the speed of light, and λ is the wavelength of photons

[tex]E = (6.626 * 10^{-34}) * (3 * 10^8) / \lambda[/tex]

λ =  [tex](6.626 * 10^{-34}) * (3 * 10^8)/201 kJ mol^{-1}[/tex]

λ = 590nm

In order for photodissociation to occur, the photon must have enough energy to break the N−O bond in the nitrate functional group of the molecule. All of the energy of the photon is efficiently used for the photodissociation.

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What is the molarity (M) of a bleach solution containing 9.50 grams of bleach (NaOCI) in 2,000 ml of solution?​

Answers

The molarity of the bleach solution is 0.0635 M.

How to calculate the molarity (M) of a solution ?

First we need to know the number of moles of solute (in this case, NaOCI) and the volume of the solution in liters.

First, we need to convert the mass of NaOCI to moles using its molar mass:

molar mass of NaOCI = 23.0 g/mol (for Na) + 16.0 g/mol (for O) + 35.5 g/mol (for Cl) + 1.0 g/mol (for I) = 74.5 g/mol

moles of NaOCI = mass / molar mass = 9.50 g / 74.5 g/mol = 0.127 moles

Next, we need to convert the volume of the solution to liters:

volume = 2,000 ml = 2,000 / 1,000 L = 2.0 L

Now we can calculate the molarity of the solution:

Molarity (M) = moles of solute / volume of solution in liters

M = 0.127 moles / 2.0 L = 0.0635 M

Therefore, the molarity of the bleach solution is 0.0635 M.

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A compound has the empirical formula given below. C3H4O3 Which compound represents the molecular formula with a scale factor of 2? A. C6H8O6 B. C2H2O2 C. C3H4O3​

Answers

The compound which represents the molecular formula with a scale factor of 2 is [tex]C_6H_8O_6[/tex] . The correct answer is option A

The empirical formula of a compound gives the simplest whole-number ratio of the atoms present in the compound. To obtain the molecular formula, we need to know the actual number of atoms in the molecule.

To find the molecular formula with a scale factor of 2, we need to multiply the subscripts in the empirical formula by 2.

Therefore, the molecular formula with a scale factor of 2 would be:

[tex]C_3H_4O_3[/tex] × [tex]2[/tex]  = [tex]C_6H_8O_6[/tex]

So option A is the correct answer.

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The combustion of ethane (C2H6)
produces carbon dioxide and steam.

2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g)


How many moles of CO2
are produced when 5.10 mol
of ethane is burned in an excess of oxygen?

Answers

Ethane (C2H6) burns to make dioxide and steam: 2C2H6(g) + 7O2(g) (g) Ethane (C2H6) fires to give methane gas and steam: 4CO2(g) + 6H2O + 2C2H6(g) + 7O2(g) (l)

How much CO2 is created if 5.70 mol5 70 molar of ethane burns in an environment with too much oxygen?

We must determine the molecules of CO2 and CO2 that were created. According to the equation above, when 2 molecules of Online C 2 H 6 are exposed to oxygen, 4 moles of Carbon dioxide C O 2 are created.

Ethane does it emit carbon dioxide?

Oxygen gas and ethane (C2H6) react to create both water and carbon dioxide. Find the quantity of carbon dioxide created when the reaction yield is 60% when 5 mol of propane is burnt with 16 molecule of oxygen initially. 2C2H6+7O2→4CO2+6H2O.

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(a) A compound X contains carbon, 66.7% and hydrogen 11.1% the rest being oxygen. Determine the empirical formula of X. (b) The vapor density of X is 36; determine the molecular formula of X.

Answers

Answer:

The molecular formula of X = C4 h8 O

Calculate the volume of the gas in cylinder B at STP.
T is 0 °C and Standard Pressure = 1.00
(standard
atm)

Answers

With one of those pieces of information, we could use the Ideal Gas Law (PV = nRT) to calculate the volume of the gas at STP.

What is STP?

STP stands for Standard Temperature and Pressure. It is defined as a temperature of 0 °C (273.15 K) and a pressure of 1 atmosphere (atm). This is the standard set of conditions used for comparing and measuring gases. At STP, 1 mole of any gas occupies a volume of 22.4 liters.

We need additional information to calculate the volume of the gas in cylinder B at STP. We would need to know either the pressure, volume, and temperature of the gas in cylinder B at its current state or the number of moles of gas in cylinder B.

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Arrange the following compounds in order of increasing percentage of oxygen by mass

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The compounds arranged in order of increasing percentage of oxygen by mass are C2H4O2, NO2, CO2, H2O, and O2. Hence, the correct option is C2H4O2, NO2, CO2, H2O, and O2.

Steps

The percentage of oxygen by mass in each compound can be calculated using their molecular weight and the weight of the oxygen atoms present in the molecule.

C2H4O2: Molecular weight = 60.05 g/mol, 4 oxygen atoms present, so %O = (4 x 16.00 g/mol) / 60.05 g/mol x 100% = 42.63%

NO2: Molecular weight = 46.01 g/mol, 2 oxygen atoms present, so %O = (2 x 16.00 g/mol) / 46.01 g/mol x 100% = 69.57%

CO2: Molecular weight = 44.01 g/mol, 2 oxygen atoms present, so %O = (2 x 16.00 g/mol) / 44.01 g/mol x 100% = 72.73%

O2: Molecular weight = 32.00 g/mol, 2 oxygen atoms present, so %O = (2 x 16.00 g/mol) / 32.00 g/mol x 100% = 100.00%

H2O: Molecular weight = 18.02 g/mol, 1 oxygen atom present, so %O = (1 x 16.00 g/mol) / 18.02 g/mol x 100% = 88.86%

Therefore, the compounds arranged in order of increasing percentage of oxygen by mass are C2H4O2, NO2, CO2, H2O, and O2. Hence, the correct option is C2H4O2, NO2, CO2, H2O, and O2.

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What is the Molarity of a 0.5 Liter solution that contains 3.84 moles of NaCI?


SHOW WORK PLEASE​

Answers

The molarity of a 0.5 liter solution that contains 3.84 moles of NaCl would be 7.68 M.

What is molarity?

Molarity is simply the ratio of the number of moles of solutes per liter volume of a solution.

In other words, molarity is a measure of the concentration of a solute in a solution. It is commonly denoted by the symbol "M" and expressed in units of mol/L.

Molarity = mole/volume

Where the volume is in liters.

In this case:

Number of moles = 3.84 molesVolume = 0.5 liter

Molarity of the NaCl solution = 3.84/0.5

                                                = 7.68 M

Thus, the molarity of a 0.5 Liter solution that contains 3.84 moles of NaCI is 7.68 M.

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NH3 is a weak base that reacts with water according to the chemical equilibrium represented above. The table provides some information for two NH3 (aq) solutions of different concentration at 25 C. Which of the following is true about the more concentrated 0.30 M NH3 (aq) and why?

Answers

The query doesn't include the chemical balance or table you mentioned. In order for me to assist you in finding the answer to your inquiry, kindly supply more details or enclose the necessary chemical equation and table.

What happens when NH3 and water react in equilibrium?

Equilibrium constant for the reaction between NH3 and water. Ammonia functions as a base when dissolved in water. Ammonium and hydroxide ions are produced after it takes hydrogen ions from water.

What causes NH3 to be a weak base in water?

Although ammonia does not naturally include hydroxide ions, when it is dissolved in water, it absorbs hydrogen ions from the water and produces both hydroxide and ammonium ions.

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Which of the following California birds adapts its bill shape based on where it lives and what it eats?
Red-Tailed Hawk
Hummingbird
Woodpecker
Jays

Answers

An external anatomical feature called the beak, bill, or rostrum is usually present in birds but can also be seen in turtles, non-avian dinosaurs, a few mammals, and other animals.

Is the beak of a woodpecker red?

These are frequent birds in parks, forests, cities, and rural settings all over the eastern and southeastern United States. Red can be seen on the rear of the head and at the base of the bill in female red-bellied woodpeckers.

Which bird's beak is red, please?

The parrot is a lovely bird. It has long green feathered tail, lovely green wings, and silky green feathers covering its body. The ruby beak of the parrot is its most alluring characteristic.

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Acid-Base Titrations When methylammonium chloride (pKa = 10.645) is titrated with tetramethylammonium hydroxide, the titration reaction is CH3NH + OH - CH3NH2 + H2O BH From B Weak acid (CH3)4N OH Weak base Find the equivalence volume in the titration of 25.0 mL of 0.010 0 M methylammonium chloride with 0.050 0 M tetramethylammonium hydroxide. Calculate the pH at Vo=0, 2.50, 5.00, and 10.00 mL. Sketch the titration curve.

Answers

5.00mL the equivalence volume in the titration of 25.0 mL of 0.010 0 M methylammonium chloride with 0.050 0 M tetramethylammonium hydroxide.

To find the equivalence volume, we need to first determine the moles of methylammonium chloride (MAC) present in the initial solution:

moles of MAC = (25.0 mL)(0.0100 M) = 0.250 mmol

Since the stoichiometry of the reaction is 1:1 between MAC and tetramethylammonium hydroxide (TMAH), the equivalence point occurs when 0.250 mmol of TMAH has been added:

moles of TMAH = (0.250 mmol)/(1000 mL/L) = 0.000250 mol

Now we can calculate the volume of TMAH required to reach the equivalence point:

0.000250 mol / 0.050 0 M = 5.00 mL

Therefore, the equivalence volume is 5.00 mL.

To calculate the pH at different points in the titration, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

At Vo=0, we have only MAC present, so [A-] = 0 and [HA] = 0.0100 M. Therefore:

pH = 10.645 + log(0.000/0.0100) = 10.645

At Vo=2.50 mL, we have added some TMAH, but have not yet reached the equivalence point. We can calculate the concentrations of the acid and conjugate base using the stoichiometry of the reaction and the amount of TMAH added:

moles of TMAH added = (2.50 mL)(0.050 0 M) = 0.000125 mol

moles of MAC remaining = 0.250 mmol - 0.000125 mol = 0.249875 mol

moles of MAC converted to MAC+ = 0.000125 mol

moles of MAC+ = 0.249875 mol

Now we can calculate the concentrations of the acid and conjugate base:

[HA] = (0.249875 mol / 0.0250 L) = 0.009995 M

[A-] = (0.000125 mol / 0.0275 L) = 0.00455 M

Using the Henderson-Hasselbalch equation:

pH = 10.645 + log(0.00455/0.009995) = 10.073

Similarly, we can calculate the pH at Vo=5.00 mL and Vo=10.00 mL:

At Vo=5.00 mL:

moles of TMAH added = (5.00 mL)(0.050 0 M) = 0.000250 mol

moles of MAC remaining = 0.250 mmol - 0.000250 mol = 0.249750 mol

moles of MAC converted to MAC+ = 0.000250 mol

moles of MAC+ = 0.249750 mol

[HA] = (0.249750 mol / 0.0250 L) = 0.009990 M

[A-] = (0.000250 mol / 0.0300 L) = 0.00833 M

pH = 10.645 + log(0.00833/0.009990) = 9.899

At Vo=10.00 mL:

moles of TMAH added = (10.00 mL)(0.050 0 M) = 0.000500 mol

moles of MAC remaining = 0.250 mmol - 0.000500 mol = 0.

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Identify and describe the role of water in the formation of landforms.

Answers

Answer:

Water participates both in the dissolution and formation of Earth's materials. The downward flow of water, both in liquid and solid form, shapes landscapes through the erosion, transport, and deposition of sediment. Shoreline waves in the ocean and lakes are powerful agents of erosion.

Below is the titration curve of a 50.0 mL of 0.1 M triprotic acid H3A with 0.1 M NaOH(aq). The initial pH and the equivalence points are indicated in the curve.
Calculate pKa1, pKa2, and pKa3 for the triprotic acid H3A.

Answers

In the titration curve, the initial pH of the triprotic acid H3A is given as approximately 1.8. This corresponds to the pH at which only H3A is present in solution.

The first equivalence point, where half of the H3A has been neutralized, occurs at approximately pH 4.3. This corresponds to the point where H2A^- is formed; therefore, Ka1 can be calculated as follows:

Ka1 = [H2A^-][H3O^+]/[H3A] = 10^(-pH1)

where pH1 is the pH at the first equivalence point.

Substituting the given values, we get:

Ka1 = [H2A^-][H3O^+]/[H3A] = 10^(-4.3)

At the second equivalence point, where H2A^- has been completely neutralized and HA^2- is formed, the pH is approximately 8.5. Thus, Ka2 can be calculated as follows:

Ka2 = [HA^2-][H3O^+]/[H2A^-] = 10^(-pH2)

where pH2 is the pH at the second equivalence point.

Substituting the given values, we get:

Ka2 = [HA^2-][H3O^+]/[H2A^-] = 10^(-8.5)

Lastly, at the third equivalence point, where HA^2- has been completely neutralized and A^3- is formed, the pH is approximately 12.3. Therefore, Ka3 can be calculated as follows:

Ka3 = [A^3-][H3O^+]/[HA^2-] = 10^(-pH3)

where pH3 is the pH at the third equivalence point.

Substituting the given values, we get:

Ka3 = [A^3-][H3O^+]/[HA^2-] = 10^(-12.3)

Thus, the values of pKa1, pKa2, and pKa3 for the triprotic acid H3A are:

pKa1 = -log(Ka1) = -log(10^(-4.3)) = 4.3
pKa2 = -log(Ka2) = -log(10^(-8.5)) = 8.5
pKa3 = -log(Ka3) = -log(10^(-12.3)) = 12.3


an alpha particle (6.70×10-27 kg) moving at a speed of 1.52×107 m/s (34000000 miles per hour).
whats the m?

Answers

According to the given statement the alpha particle's wavelength is 6.48×10⁻¹⁵ meters.

What is alpha particle made of?

The positively charged alpha particles (α) are created when the nucleus of an atom's two protons as well as two neutrons join forces. Alpha particles are created by the radioactivity of the most dangerous radioactive elements, including uranium, radium, and polonium.

The de Broglie wavelength equation may be used to determine the alpha particle's wavelength:

λ = h / p

where p is the particle's momentum, h is its Planck constant (6.626 * 10⁻³⁴ J s), and is its wavelength.

The mass (m) and velocity (v) of an alpha particle may be used to compute its momentum:

p = m * v

Inputting the values provided yields:

p = (6.70×10⁻²⁷ kg) * (1.52×10⁷ m/s) = 1.02×10⁻¹⁹ kg m/s

Now we can determine the wavelength using this value:

λ = h / p = 6.626 x 10⁻³⁴ J s / 1.02×10⁻¹⁹ kg m/s

λ = 6.48 x 10⁻¹⁵ m

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The complete question is -

Calculate the wavelength of -

An alpha particle (6.70×10⁻²⁷ kg) moving at a speed of 1.52×10⁷ m/s (34000000 miles per hour). What is the m ?

3. What is the molarity of a solution that contains 0.50 moles of NaOH dissolved in 500 mL solution? ​

Answers

Answer:

1M

Explanation:

To find molarity, we use the formula [tex]\frac{n}{v}[/tex], where n is the moles of solute, and v is the volume in liters.

This formula is fairly straightforward. When we input our values we get a formula of [tex]\frac{0.50 moles}{0.5 liters}[/tex]. Simplifying yields 1M. (Hint: Don't forget that occasionally we do have to convert between units. In this case, the 500mL was converted to liters using the known conversion factor that 1000mL= 1 liter.)

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