please help me with this lab i wasn’t here for!

3. now that you have the mass of the NaHCO3 reactant, and the mass of the product NaCI , convert each to moles and compare to the mole ratio from your balanced equation C space below for your calculations

mass NaHCO3:
mass NaCI:
moles NaHCO3:
moles NaCI:

does the mole to mole ratio for your reaction? Agree with the ratio for the balanced equation?___

4. which reactant is the excess reactant for your reaction, how do you know?

5. Using the limiting reactant calculate the maximum amount of product that can be made from this reaction.

6. using the theoretical yield in the mass of the product that you put produce calculate percent yield.

calculations:
question #3: converting mass to moles

question #5: calculating the theoretical yield

question #6: calculating percent yield

Answer:

Explanation:

[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.

After that the molar ratio between HgO and O must be found but in this case its 1:1

[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out  with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.

[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen

The precipitation of an ionic substance from solution occurs when the ionic product exceeds the value of its solubility product at that temperature. Here the moles of Na₂S₂O₃ needed is

The solubility product of a sparingly soluble salt is defined as the product of the molar concentrations of its ions in a saturated solution of it at a given temperature.

Here the concentration of Ag⁺ ions = √Ksp = √3.3 × 10⁻¹³ = 1.81 × 10⁻¹³.

Moles of Ag⁺ ions: (1.82 x 10⁻¹³ M) x 1.0 L = 1.82 x 10⁻¹³  mol Ag⁺

Use the stoichiometry of the reaction to find the moles of Na₂S₂O₃ needed: 1 mol Na₂S₂O₃ / 2 mol Ag⁺ = 0.5 mol Na₂S₂O₃/mol Ag⁺

Moles of Na₂S₂O₃ required: 0.5 mol Na₂S₂O₃/mol Ag⁺ x 1.82 x 10⁻¹³ mol Ag⁺ = 9.1 x 10⁻¹⁴ mol Na₂S₂O₃

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CH₃CHCl₂COOH is 2,2-dichloropropanoic acid, with the least pH, option (c) is correct.

pH is a measure of the acidity or basicity of a solution. A lower pH indicates a higher acidity. Acidity is due to the presence of hydrogen ions (H⁺) in a solution. The more the concentration of H⁺, the lower the pH. CH₃CH₂COOH is propanoic acid, which has a pH of around 4.9.

CH₂ClCH₂COOH is 2-chloropropanoic acid, which has a pH of around 2.8 due to the electron-withdrawing effect of the chlorine atom. CH₃CH₂CH₂COOH is butanoic acid, which has a pH of around 4.8. Thus, CH₃CHCl₂COOH is 2,2-dichloropropanoic acid, which has the least pH among the given options, around 1.5 due to the presence of two electron-withdrawing chlorine atoms, option (c) is correct.

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The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation, we get:

n = PV/RT

For the container of helium:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Now, using the same equation for the container of xenon:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Therefore, there are also 0.0688 moles of xenon gas present in the container.

The volume of N2 produced each day is approximately 24.56 L.

First, let's find the number of moles of NO3- that flow into the swamp each day:

NO3- molar mass = 14.01 (N) + 3 * 16.00 (O) = 62.01 g/mol

135 g / 62.01 g/mol ≈ 2.177 moles of NO3-

In the denitrification process, NO3- is reduced to N2 gas. The balanced equation for denitrification is:

2NO3- → N2 + 3O2

From the stoichiometry of the reaction, we can see that 2 moles of NO3- produce 1 mole of N2. Therefore, the moles of N2 produced each day can be calculated as follows:

moles of N2 = 2.177 moles of NO3- / 2 ≈ 1.0885 moles of N2

Now we can use the ideal gas law equation to find the volume of N2 produced:

PV = nRT

where:

P = Pressure (1.00 atm)

V = Volume (in Liters)

n = Moles of N2 (1.0885 moles)

R = Ideal gas constant (0.0821 Latm/molK)

T = Temperature (17.0°C or 290.15 K)

Solving for the volume of N2:

V = nRT / P

V = (1.0885 moles) * (0.0821 Latm/molK) * (290.15 K) / (1.00 atm)

V ≈ 24.56 L

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Answer:

D

Explanation:

to keep seeds safe and make them easier to spread

Answer:

D. to keep seeds safe and make them easier to spread

Explanation:

The main reason plants grow fruit is to aid in the protection and spreading of seeds. The fruit protects the seeds and also helps to spread them. Many fruits are good to eat and attract small animals, such as birds and squirrels, who like to feed on them. The seeds pass through them unharmed and then get spread through their droppings. So, the correct answer would be D.

Answer:

70.91 kJ

Explanation:

The amount of energy (in kJ) required to increase the temperature of 255 g of water from 25.2 C to 90.5 C can be calculated using the formula:

Q = m * c * ΔT

Where Q is the amount of energy, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

Substituting the given values:

m = 255 g

c = 4.184 J/g C

ΔT = (90.5 - 25.2) C = 65.3 C

Q = 255 g * 4.184 J/g C * 65.3 C

Q = 70905.564 J

Q = 70.91 kJ (rounded to two decimal places)

Therefore, the amount of energy required to increase the temperature of 255 g of water from 25.2 C to 90.5 C is 70.91 kJ.

There are approximately 223.2 grams of oxygen in 615 grams of N2O.

To find the number of grams of O in 615g of N2O, we first need to understand the chemical formula of N2O. N2O is a compound made up of two nitrogen atoms (N) and one oxygen atom (O). Therefore, the molecular weight of N2O would be:
(2 x atomic weight of N) + (1 x atomic weight of O)
= (2 x 14.01 g/mol) + (1 x 16.00 g/mol)
= 44.01 g/mol
Now, to calculate the number of grams of O in 615g of N2O, we need to know the proportion of O in the compound. Since there is only one oxygen atom in each molecule of N2O, we can find the proportion of O by dividing the atomic weight of O by the molecular weight of N2O:
Atomic weight of O / Molecular weight of N2O
= 16.00 g/mol / 44.01 g/mol
= 0.363
This means that oxygen makes up 36.3% of the total weight of N2O. To find the number of grams of O in 615g of N2O, we can multiply the total weight by the proportion of O:
615g x 0.363
= 223.2g

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PbS, sometimes referred to as galena, is made up of lead (Pb) and sulfur (S) atoms and has a straightforward structure. Each lead atom forms a tetrahedral link with four sulfur atoms, while each sulfur atom forms a covalent bond with two lead atoms.

Each sulfur atom forms an octahedral link with six iron atoms, while each iron atom forms a covalent bond with two sulfur atoms. FeS2's crystal structure is a cubic, tightly packed lattice.

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The pH of the solution is 5.72, which is slightly acidic.

pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the hydrogen ion concentration (H+) in a solution. The pH scale ranges from 0 to 14, where a pH of 7 is neutral, pH below 7 is acidic, and pH above 7 is basic. The formula to calculate pH is pH = -log[H+], where [H+] represents the concentration of hydrogen ions in moles per liter.
Given the H+ concentration of 1.9x10-6, we can calculate the pH of the solution as follows:
pH = -log(1.9x10-6) = 5.72
It is important to note that pH is an important factor in various chemical and biological processes. It can affect the solubility of certain substances, enzymatic activity, and the growth and survival of living organisms. Maintaining the appropriate pH is crucial for the proper functioning of these processes.

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a) We need 32.6 liters of [tex]O_2[/tex] at 41 °C and 0.307 atm to burn 8.57 L of [tex]C_2H_6[/tex]  at 41 °C and 0.307 atm

b) 18.5 liters of [tex]CO_2[/tex] will be produced at 41 °C and 0.307 atm.

To answer this question, we will use the ideal gas law, which relates pressure, volume, temperature, and number of moles of a gas. We will also use stoichiometry to relate the amount of ethane and oxygen consumed and the amount of carbon dioxide and water produced.

(a) To determine how many liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex] , we first need to convert the volume of ethane to moles using the ideal gas law:
n([tex]C_2H_6[/tex] ) = PV/RT = (0.307 atm)(8.57 L)/(0.0821 L·atm/mol·K)(314 K) = 0.342 mol

From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] react with 7 moles of [tex]O_2[/tex] . Therefore, the amount of [tex]O_2[/tex] needed is:
n([tex]O_2[/tex]) = (7/2) n([tex]C_2H_6[/tex]) = (7/2)(0.342 mol) = 1.20 mol

Now we can use the ideal gas law again to calculate the volume of [tex]O_2[/tex] needed:
V([tex]O_2[/tex] ) = n([tex]O_2[/tex])RT/P = (1.20 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 32.6 L

Therefore, 32.6 liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex]  at at 41 °C and 0.307 atm

(b) From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] produce 4 moles of [tex]CO_2[/tex] . Therefore, the amount of [tex]CO_2[/tex] produced is:

n([tex]CO_2[/tex]) = 2 n([tex]C_2H_6[/tex]) = 2(0.342 mol) = 0.684 mol

V([tex]CO_2[/tex]) = n([tex]CO_2[/tex])RT/P = (0.684 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 18.5 L

Therefore, 18.5 liters of [tex]CO_2[/tex] at 41 °C and 0.307 atm will be produced.

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Quantifying chemical reactions is essential in understanding the stoichiometry of a reaction, predicting product formation, and optimizing product yield in industrial applications. Stoichiometric coefficients and limiting reactants are two important tools used in this process.

Quantifying chemical reactions involves measuring the amount of reactants and products involved in a chemical reaction. This is important in determining the stoichiometry of the reaction, which refers to the relative amounts of reactants and products involved. Stoichiometry is a crucial concept in chemistry because it allows scientists to predict the amount of product that will be formed from a given amount of reactant, or vice versa.
One way to quantify chemical reactions is through the use of stoichiometric coefficients. These coefficients represent the number of moles of each reactant and product involved in the reaction. For example, the balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water is:
[tex]2H2 + O2 → 2H2O[/tex]
This equation tells us that two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water. The stoichiometric coefficients can be used to determine the mass of each reactant and product involved in the reaction, using the molar masses of each substance.
Another way to quantify chemical reactions is through the use of limiting reactants. A limiting reactant is the reactant that is completely consumed in a reaction, limiting the amount of product that can be formed. The amount of product formed will be determined by the amount of limiting reactant present. This concept is important in industrial chemistry, where maximizing product yield is often the goal.

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Solutions of [tex]Pb(NO_3)^2[/tex] and [tex]NaCl[/tex]  are combined, resulting in concentration of 0.0050 M [tex]Pb(NO_3)^2[/tex] and 0.0025 M [tex]NaCl[/tex] immediately upon mixing. The correct description of the final solution, given that the Ksp of [tex]PbCl_2[/tex] is 1.70×10^-5 is All solutes remain soluble. The correct answer is option A

Upon mixing [tex]Pb(NO_3)^2[/tex] and [tex]NaCl[/tex] , the following reaction occurs:

[tex]Pb(NO_3)^2[/tex] + [tex]2NaCl[/tex]  → [tex]PbCl_2[/tex] +[tex]2NaNO_3[/tex]

Using the given concentrations of the reactants, the reaction quotient Qc can be calculated as:

Qc =[tex][Pb^2^+][Cl^-]^2[/tex] = [tex](0.0050 M)(0.0025 M)^2[/tex]  

Qc [tex]= 3.13[/tex] × [tex]10^{-3}[/tex]

Comparing Qc to the solubility product constant (Ksp) of [tex]PbCl_2[/tex] , we see that Qc < Ksp. This indicates that the system is not at equilibrium and more [tex]PbCl_2[/tex] can dissolve before the product reaches saturation.

Therefore, no precipitation of [tex]PbCl_2[/tex] will occur, and option A is the correct answer: all solutes remain soluble.

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Answer:

Explanation:

It seems like you’re trying to count the number of atoms in some chemical formulas. Here’s the list of the number of each atom in the formulas you provided:

Formula 1: Н - 1 Formula 2: H - 1, O - 1 Formula 3: Н - 14 Formula 4: Н - 2, C - 3 Formula 5: I - 1 Formula 6: Н - 3 Formula 7: H - 2 Formula 8: Н - 2, C - 3, O - 1 Formula 9: Li - 1 Formula 10: Н - 2 Formula 11: Н - 2 Formula 12: H - 1 Formula 13: Н - 2, C - 2, O - 1 Formula 14: II

The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

we can use the principles of chemical equilibrium and the mole fraction formula.

First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:

A + 2B <=> C + D

where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.

Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:

Kc = [C][D] / [A][B]²

where [X] denotes the molar concentration of species X at equilibrium.

Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:

Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol

We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:

[C] = 0.9 mol / 4 L = 0.225 M

Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:

Kc = [C][D] / [A][B]²

0.9 = (0.225)(D) / (1)(2²)

D = 1.8

Therefore, the molar concentration of D at equilibrium is 1.8 M.

Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:

[A] = 1 mol / 4 L = 0.25 M

[B] = 2 mol / 4 L = 0.5 M

Mole fraction of X = moles of X / total moles

Mole fraction of A = 1 mol / 4 mol = 0.25

Mole fraction of B = 2 mol / 4 mol = 0.5

Mole fraction of C = 0.9 mol / 4 mol = 0.225

Mole fraction of D = 1 mol / 4 mol = 0.25

Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

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Answer:

Cvapor = 2.03 J/(g·°C)heu

The electrons can be arranged in increasing order of energy as follows: (1,0,0,-1/2) < (2,1,0,-1/2) < (2,1,0,-1/2) < (3,1,1,1/2) < (3,2,0,-1/2).

The energy of an electron is determined by its principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m), and spin quantum number (s). The electrons can be arranged in increasing order of energy by comparing their quantum numbers.

Starting with the lowest energy electron, we have the electron with quantum numbers (1,0,0,-1/2). This electron has the lowest principal quantum number, indicating that it occupies the lowest energy level.

It also has an azimuthal quantum number of zero, which corresponds to the s subshell, and a negative spin quantum number, indicating that its spin is aligned opposite to the magnetic field.

Next, we have the two electrons with quantum numbers (2,1,0,-1/2). These electrons have the same principal quantum number, indicating that they occupy the same energy level.

They both have an azimuthal quantum number of one, which corresponds to the p subshell, and a negative spin quantum number.

Following these electrons, we have the electron with quantum numbers (3,1,1,1/2). This electron has a higher principal quantum number than the previous electrons, indicating that it occupies a higher energy level.

It has an azimuthal quantum number of one, which corresponds to the p subshell, and a positive spin quantum number.

Finally, we have the electron with quantum numbers (3,2,0,-1/2). This electron has the highest azimuthal quantum number of all the electrons, indicating that it occupies the d subshell. It also has a negative spin quantum number.

Therefore, the electrons can be arranged in increasing order of energy as follows: (1,0,0,-1/2) < (2,1,0,-1/2) < (2,1,0,-1/2) < (3,1,1,1/2) < (3,2,0,-1/2).

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Answer:

B

self explanatory

Explanation:

The volume of the 1.5M solution of HCl solution that has 145g of mass is 2.65L.

The volume of a solution can be calculated by dividing the number of moles by its molar concentration as follows;

Volume = no of moles ÷ molarity

According to this question, a 1.5M solution of HCl has 145g of mass. The number of moles can be calculated as follows:

no of moles = 145g ÷ 36.5g/mol = 3.973 moles

volume of HCl solution = 3.973mol ÷ 1.5M

volume of HCl = 2.65L

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At equivalence point, the reaction is seen to consume approximately 0.0024973 moles of KHP and then 0.0024973 moles of  NaOH

The primary aim in this problem is to standardize a solution of the sodium hydroxide, NaOH, with the aid of potassium hydrogen phthalate, KHP.

The beginning point in this problem is the balanced chemical equation with respect to this neutralization reaction

KHP (aq] + NaOH(aq] → KNaP(aq] + H₂O(l]

The important thing that we are going to observe is that there is a ratio of 1:1 mole ratio between the two reactants. This suggests to us that the equivalence point can be attained by getting equal number of moles of  KHP and of NaOH to react with each other.

We will begin with 0.5100 g of KHP. To obtain the molar amount of acid utilized for the experiment, we will make use of its molar mass of 0.5100g⋅

molar mass of KHP

1 mole KHP 204.22g = 0.0024973 moles KHP

Thus, at equivalence point, the reaction is seen to consume approximately 0.0024973 moles of KHP and then 0.0024973 moles of  NaOH, due to the fact that it's what the  1:1 mole ratio suggests to us.

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[tex]FeCl_3[/tex] is added to ensure all reducing agent is oxidized, indicating endpoint in dichromate titration.

In a dichromate titration,   [tex]FeCl_3[/tex] is frequently added to the example arrangement as a sign of the endpoint. The expansion of [tex]FeCl_3[/tex] to the example arrangement assists with guaranteeing that the lessening specialist has been all oxidized by the potassium dichromate ([tex]K_2Cr_2O_7[/tex] ) arrangement.

[tex]FeCl_3[/tex] responds with any overabundance[tex]K_2Cr_2O_7[/tex]  in the answer for structure a red-earthy colored encourage of[tex]Fe(OH)_3[/tex] , demonstrating that the lessening specialist has been all oxidized. This response is known as a "back-titration" since overabundance[tex]K_2Cr_2O_7[/tex]   is added to the arrangement, trailed by the option of [tex]FeCl_3[/tex] to decide how much unreacted [tex]K_2Cr_2O_7[/tex] .

The response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] can be addressed as:

[tex]6FeCl_3 + K_2Cr_2O_7 + 7H_2SO_4 → 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O + 3Cl_2[/tex]

The [tex]FeCl_3[/tex] goes about as a pointer in this response since it responds with the overabundance [tex]K_2Cr_2O_7[/tex] until the decreasing specialist has been all oxidized. As of now, the red-earthy colored encourage of [tex]Fe(OH)_3[/tex]  structures, showing that the endpoint has been reached.

Without the expansion of [tex]FeCl_3[/tex], it would be challenging to precisely decide the endpoint of the titration. The expansion of [tex]FeCl_3[/tex] is important to guarantee that the lessening specialist has been all oxidized and that the endpoint has been reached, considering a more exact assurance of the grouping of the diminishing specialist in the example arrangement.

In outline, the expansion of [tex]FeCl_3[/tex] to the example arrangement in a dichromate titration is significant in light of the fact that it assists with guaranteeing that the decreasing specialist has been all oxidized, considering a more precise assurance of its focus.

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The complete question is -

What is the reason for a dichromate titration, and how does [tex]FeCl_3[/tex]support deciding the endpoint of the titration? Might you at any point give the compound condition to the response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] and make sense of why [tex]FeCl_3[/tex] goes about as a marker in this response?

(NH4)3PO4 is insoluble in water. The correct option is D

According to their chemical formula and ionic charges, ionic compounds generally follow a set of solubility laws that define their solubility patterns in water. These guidelines aid in determining whether an ionic compound will dissolve in water or not as well as if it will precipitate when combined with other ionic compounds.

Therefore, (NH4)3PO4 is the compound that is expected to be insoluble in water based on the solubility rules.

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Final temperature: 677.4K. Work done: -7026J.

Heat exchanged: 0J. Change in internal energy: -7026J.

(i) For an adiabatic process, T1(V1)^γ-1 = T2(V2)^γ-1.

When we substitute the values (γ=1.4, T1=300K, V1=6dm³, V2=2dm³), we get T2 = 677.4K.

(ii) w = -(P1V1 - P2V2)/(γ-1) = -(nRT1 - nRT2)/(γ-1) = -5 * 8.314 * (677.4 - 300) / 0.4 = -7026J.

For adiabatic, q = 0. ΔE = q + w = -7026J (since q=0).

Final temperature: 677.4K. Work done: -7026J.

Heat exchanged: 0J. Change in internal energy: -7026J.

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Answer:

No,a pH of 5 is slightly acidic,not neutral. A pH of 7 is considered neutral

Answer:O has 6, Na has 1, and Sr has 2.

Explanation:

A chemical interaction between an acid and a base is known as an acid-base reaction.

Thus, These are known as acid-base theories, such as the Brnsted-Lowry acid-base theory, and they offer alternative conceptions of the reaction mechanisms and their application in solving related problems.

When examining acid-base reactions for gaseous or liquid species, or when the acid or basic character may be less obvious, their significance becomes clear.

The relative potency of the conjugated acid-base pair in the salt controls the pH of its solutions when weak acids and bases react. The resulting salt or its solution can be basic, neutral, or acidic. A strong acid and a weak base can combine to generate an acid salt.

Thus, A chemical interaction between an acid and a base is known as an acid-base reaction.

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i) The equation for the reaction between hydrochloric acid and zinc is:

[tex]Zn + 2HCl → ZnCl2 + H2[/tex]

ii) n(HCl) = C × V = 1.0M × 0.05 L = 0.05 moles

iii) The volume of gas evolved at STP is 0.544 L or 544 mL.

The concentration of hydrochloric acid is 1.0M, which means that there is 1 mole of hydrochloric acid in 1 liter (1000 cm³) of solution. The volume of the hydrochloric acid used is 50 cm³, which is 0.05 liters.

According to the stoichiometry of the reaction, each mole of hydrochloric acid produces one mole of hydrogen ions, so the number of moles of hydrogen ions in the solution is also 0.05 moles.

The volume of gas evolved can be calculated from the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature of the gas in Kelvin. At standard temperature and pressure (STP), the pressure is 1 atm and the temperature is 273 K. The molar volume of a gas at STP is 22.4 L/mol.

From the equation for the reaction, we know that one mole of hydrogen gas is produced for every two moles of hydrochloric acid used. Therefore, the number of moles of hydrogen gas produced is:

n(H2) = 0.5 × n(HCl) = 0.5 × 0.05 moles = 0.025 moles

Using the ideal gas law, we can calculate the volume of hydrogen gas produced at STP:

V(H2) = n(H2) × RT/P = 0.025 mol × 0.0821 L·atm/K·mol × 273 K/1 atm = 0.544 L

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There are 3 bonding pairs and 7 lone pairs of electrons in N2.

An atom's nucleus is orbited by an electron, a subatomic particle with a negative charge. Along with protons and neutrons, it is one of the elementary particles that make up matter. The mass of an electron is exceedingly small, it is roughly 1/1836 that of a proton.

To determine the number of lone pairs of electrons in N2, we need to subtract the number of bonding pairs from the total number of valence electrons:

Number of lone pairs = Total number of valence electrons - Number of bonding pairs

Number of lone pairs = 10 - 3

Number of lone pairs = 7

Therefore, there are 3 bonding pairs and 7 lone pairs of electrons in N2.

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The difference in Pa between the pressure in the bag and the atmospheric pressure is 1.505 kPa.

To obtain the difference in pressure, we first need to know the atmospheric pressure near sea level. This is 760 mm Hg. When we convert this to pascals, we will have,  101.32472 kPa.

Now, the difference in pressure will be obtained by subtracting the atmospheric pressure in the bag from the atmospheric pressure near sea level and this is:

101.32472 kPa -  99.82 kPa

= 1.505 kPa.

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The pressure of the gas in the smaller flask at the new temperature is approximately 168.5 mm Hg.

To solve this problem, we can use the combined gas law equation, which relates the initial and final states of a gas sample undergoing changes in pressure, volume, and temperature. The equation is:
[tex]P_1V_1/T_1 = P_2V_2/T_2[/tex]

where [tex]P_1[/tex] and [tex]P_2[/tex] are the initial pressure and final pressure, [tex]V_1[/tex] and [tex]V_2[/tex] are the initial and final volumes, and [tex]T_1[/tex] and [tex]T_2[/tex] are the initial and final temperatures in Kelvin.

[tex]V_1[/tex] = 245 mL
[tex]T_1[/tex] = 23.5°C + 273.15 = 296.65 K
[tex]P_1[/tex] = 37.8 mm Hg
[tex]V_2[/tex] = 54 mL
[tex]T_2[/tex] = 0°C (ice water) + 273.15 = 273.15 K

We need to find [tex]P_2[/tex] . Plug the given values into the equation and solve for [tex]P_2[/tex] :
(37.8 mm Hg * 245 mL) / 296.65 K = (P2 * 54 mL) / 273.15 K

Rearrange the equation to isolate [tex]P_2[/tex] :
[tex]P_2[/tex] = (37.8 mm Hg * 245 mL * 273.15 K) / (296.65 K * 54 mL)
[tex]P_2[/tex] ≈ 168.5 mm Hg
So, the pressure of the gas is approximately 168.5 mm Hg in the smaller flask at the new temperature.

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