What is the molar mass of a compound if a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr? The temperature in Celsius is known to two significant figures.

The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium bicarbonate [tex](NaHCO_3)[/tex] is:

[tex]HCl + NaHCO_3\ - > NaCl + CO_2 + H_2O[/tex]

The coefficients in the balanced equation show that 1 mole of HCl reacts with 1 mole of [tex]NaHCO_3[/tex] to produce 1 mole of NaCl, 1 mole of [tex]CO_2[/tex], and 1 mole of [tex]H_2O[/tex]. We need to find the number of moles of [tex](NaHCO_3)[/tex] present in the tablet.

2 grams of [tex]NaHCO_3[/tex] is equivalent to 0.02 moles, and 18 grams of HCl is equivalent to 0.45 moles. Since [tex](NaHCO_3)[/tex] is limiting reagent, only 0.02 moles of NaCl and [tex]CO_2[/tex] will be produced. The molar mass of [tex]CO_2[/tex] is 44 g/mol, so the mass of [tex]CO_2[/tex] produced is 0.88 g. The molar mass of NaCl is 58.44 g/mol, mass of NaCl produced is 1.17 g.

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The nucleus completing the following equation is option C: ₂₄⁵⁰Cr.

This reaction is a type of radioactive nuclei decay.

Radioactive decay is the process by which unstable atomic nuclei undergo spontaneous transformations in order to achieve a more stable state. This is accomplished by the emission of particles and/or electromagnetic radiation from the nucleus. The decay may occur by several mechanisms, including alpha decay, beta decay, gamma decay, and electron capture.

In alpha decay, the nucleus emits an alpha particle, which consists of two protons and two neutrons, resulting in a daughter nucleus that has two fewer protons and two fewer neutrons than the original nucleus.

In beta decay, a neutron in the nucleus is converted into a proton and an electron, and the electron is then emitted from the nucleus as a beta particle. This results in the daughter nucleus having one more proton and one fewer neutron than the original nucleus.

In gamma decay, the nucleus emits a gamma ray, which is a high-energy electromagnetic radiation, without changing the number of protons or neutrons in the nucleus.

In electron capture, an electron from the inner shell of the atom is captured by the nucleus, and a proton in the nucleus is converted into a neutron. This results in the daughter nucleus having one fewer proton and one more neutron than the original nucleus.

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Answer:

Explanation:

a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:

Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)

In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.

b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.

The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.

The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol

According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.

Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.

The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.

The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.

c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.

The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.

The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.

The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.

Answer:

Vestigial

Explanation:

The retention of genetically determined traits or structures that have partially or completely lost their ancestral purpose in a specific species is known as vestigiality. In most cases, evaluating the vestigality requires comparison with comparable traits in closely related species.

Answer: C

Explanation:

Answer: C

Explanation:

Question #3: 0.8 g of NaHCO3 mass NaCI weight: 0.4 g 0.8 g/84 g/mol, or 0.0095 moles, of NaHCO3 0.4 g/58.5 g/mol = 0.0068 moles of NaCI are the moles.

The reaction's mole to mole ratio and the ratio in the balanced equation (1:1) are in agreement. The highest quantity of NaCI that may be produced from this reaction is 0.0095 moles since NaHCO3 is the limiting reactant.

The theoretical yield of NaCI is 0.0095 moles, which is question #6. The finished product weighs 0.4 g. The percent yield is 0.4 g/0.0095 moles times 100, which is 42.1%.

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Answer:

Br, Se, Sr, Rb

Explanation:

Atomic radius increases as you move to the left and down the periodic table. The increase in radius as you move left is due to decreasing effective nuclear charge (the pull an electron feels from the nucleus) since the number of protons decrease. The increase in radius as you move down is due to a higher number of principle energy levels (orbital in which the electron is located relative to the atom's nucleus), causing the electrons to be farther from the nucleus.

The ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

The relationship between ΔG and K is given by the following equation:

ΔG = -RTln(K)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 °C = 298.15 K), and ln denotes the natural logarithm.

To calculate K, we need to use the equilibrium expression and the given concentrations:

[tex]K = [C]/([A][B])[/tex]

[tex]K = (5.00 * 10^{-5} M)/((1.50 M)(1.00 M))[/tex]

[tex]K = 3.33 x 10^{-5}[/tex]

Now we can substitute the values for R, T, and K into the equation for ΔG:

ΔG = -RTln(K)

ΔG = [tex]-(8.314 J/(mol.K))(298.15 K)ln(3.33 x 10^{-5})[/tex]

ΔG = -8.35 kJ/mol

Therefore, the ΔG value for the reaction A (aq) + B (aq) → C(aq) at 25 °C and the given concentrations is -8.35 kJ/mol.

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Answer:

Explanation:

The chemical equation for the production of manganese from manganese (II) carbonate, oxygen, and aluminium can be represented as follows:

3MnCO3(s) + 3O2(g) + 4Al(s) → 3Mn(s) + 3CO2(g) + 2Al2O3(s)

In this equation, manganese (II) carbonate (MnCO3) reacts with oxygen (O2) and aluminium (Al) to produce manganese (Mn), carbon dioxide (CO2), and aluminium oxide (Al2O3). The equation is balanced with three molecules of manganese carbonate, three molecules of oxygen, and four molecules of aluminium reacting to produce three molecules of manganese, three molecules of carbon dioxide, and two molecules of aluminium oxide.

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the answer is 2.5 according to me

The complete equation for the neutralization reactions for the LiOH + HNO₂ is as :

LiOH  +  HNO₂ ---->  LiNO₂  +  H₂O

The Neutralization reaction is the reaction as in the chemical reaction in which the acid will reacts with the base and to produce the salt and the water molecule. The general equation of the chemical reaction is as :

HX  +  BOH  -->  BX  + H₂O

The reaction with the LiOH and the HNO₂ is :

LiOH  +  HNO₂ ---->  LiNO₂  +  H₂O

There is the combination of the H⁺ ions and OH⁻ ions that will form the water.

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We can use the combined gas law to determine the pressure of the gas at the final state. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 1.01 atm), volume (V1 = 2.31 L), and temperature (T1 = 279 K) of the gas, and the final volume (V2 = 1.09 L), and temperature (T2 = 308 K) of the gas. We can solve for P2, the final pressure of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

P2 = (P1 x V1 x T2) / (V2 x T1)

P2 = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P2 = 2.41 atm (rounded to three significant figures)

Therefore, the pressure of the gas when the volume is 1.09 L and the temperature is 308 K is approximately 2.41 atm.

The electronic configuration of the d-block metal ions in the 3d series is represented by electronic configuration of Argon (Ar), 3d and 4s sub orbitals.

The electronic configuration of the d-block metal ions in the 3d series is as follows:

Scandium (Sc): [Ar] 3d¹ 4s²

Titanium (Ti): [Ar] 3d² 4s²

Vanadium (V): [Ar] 3d³ 4s²

Chromium (Cr): [Ar] 3d⁵ 4s¹

Manganese (Mn): [Ar] 3d⁵ 4s²

Iron (Fe): [Ar] 3d⁶ 4s²

Cobalt (Co): [Ar] 3d⁷ 4s²

Nickel (Ni): [Ar] 3d⁸ 4s²

Copper (Cu): [Ar] 3d¹⁰ 4s¹

Zinc (Zn): [Ar] 3d¹⁰ 4s²

Thus, the above illustration shows the electronic configuration of all the metal ions in the d-blocks (3d series)​.

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Answer:

Explanation:

Both 2CH4 and C2H8 have the same number and kind of elements. But practically, 2CH4 will be existing but C2H8 cannot exist.

In the titration between HCl and NaOH, the medium is neutral at the end point because of complete neutralization of a strong acid by a strong base.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.

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Answer:

Atomic number of sodium is 11

Electronic configuration of a sodium atom :

Since sodium has one electron in its outermost shell, Therefore, sodium can easily donate it's one electron. As the result it becomes sodium cation with + 1 charge.

Electronic configuration of a sodium cation,[tex] \: \sf ({Na}^{+1}) [/tex]

In case of sodium cation, it has fully filled electronic configuration.

Cations - Atoms that carry postive charge are called cations. Cations are formed when an atom loses its electron.

For example : [tex]\sf {Na}^{+} [/tex]

Anions - Atoms that carry negative charge are called anions. Anions are formed when an atom gains a electron.

For example : [tex]\sf {Cl}^{-} [/tex]

The 210.00 g sample of the water with the initial temperature of the 29.0°C absorbs the 7,000.0 J of heat. The final temperature of the water is the 36.9  °C .

The mass of the water = 210 g

The initial temperature = 29.0 °C

The final temperature = ?

The heat energy = 7000 J

The specific heat capacity = 4.184 J/g  °C

The heat energy is expressed as :

Q = m c ΔT

Where,

The m is mass of water = 210 g

The c is specific heat of water = 4.184 J/g  °C

The  ΔT is change in temperature = final temperature - initial temperature

The  ΔT is change in temperature = T - 29.0 °C

7000 = 210 × 4.184 ( T - 29.0  )

7000 = 878.64 ( T - 29.0  )

( T - 29.0  ) = 7.966

T = 36.9  °C

The final temperature is 36.9  °C .

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To solve this problem, we can use the formula:

q = m × c × ΔT

where q is the heat absorbed or released, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.

First, let's calculate the mass of water:

m = 225.0 g

Next, let's calculate the heat absorbed by the water:

q_water = m × c × ΔT

q_water = 225.0 g × 4.184 J/(g·°C) × (24.60°C - 20.53°C)

q_water = 3749.8 J

Since the metal released 4274 J of heat, the heat absorbed by the calorimeter can be calculated by subtracting the heat absorbed by the water from the total heat released by the metal:

q_calorimeter = - (q_water + q_metal)

q_calorimeter = - (3749.8 J + 4274 J)

q_calorimeter = - 8023.8 J

Therefore, the heat absorbed by the calorimeter is -8023.8 J, which is approximately equal to -8000 J or -8.0 kJ. The answer is (c) -339 J, since it is the closest to the calculated value when rounded to the nearest integer. Note that the negative sign indicates that the calorimeter absorbed the heat, which is expected since the reaction involved a release of heat.

The theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

To calculate the theoretical yield of MgO, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO):

2Mg + O₂ → 2MgO

According to the stoichiometry of this equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we need to determine the number of moles of Mg in each trial and use the mole ratio to find the theoretical yield of MgO.

For Trial 1:

The mass of Mg used is: 26.682 g - 27.012 g = 0.330 g

The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:

0.330 g / 24.31 g/mol = 0.0136 mol Mg

According to the balanced equation, 2 moles of Mg produce 2 moles of MgO, so the theoretical yield of MgO is:

0.0136 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.348 g MgO

For Trial 2:

The mass of Mg used is: 26.987 g - 26.695 g = 0.292 g

The number of moles of Mg is:

0.292 g / 24.31 g/mol = 0.0120 mol Mg

The theoretical yield of MgO is:

0.0120 mol Mg x (2 mol MgO / 2 mol Mg) x (40.31 g MgO/mol) = 0.307 g MgO

To calculate the percent yield of MgO, we need to use the following formula:

Percent yield = (actual yield / theoretical yield) x 100%

For Trial 1:

The actual yield of MgO is: 27.214 g - 27.012 g = 0.202 g MgO

The percent yield of MgO is:

(0.202 g / 0.348 g) x 100% = 58.0%

For Trial 2:

The actual yield of MgO is: 27.183 g - 26.695 g = 0.488 g MgO

The percent yield of MgO is:

(0.488 g / 0.307 g) x 100% = 159.2%

To calculate the average percent yield of MgO for the two trials, we add the percent yields and divide by 2:

Average percent yield = (58.0% + 159.2%) / 2 = 108.6%

Therefore, the theoretical yield of MgO for Trial 1 is 0.348 g, and for Trial 2 is 0.307 g. The percent yield of MgO for Trial 1 is 58.0% and for Trial 2 is 159.2%. The average percent yield of MgO for the two trials is 108.6%.

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The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L

Molarity is  described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.

Molarity = moles of solute / liters of solution

we then substitute the given values, and have

Molarity = 2.0 moles / 3.0 L

Molarity = 0.67 mol/L

Molarity is  very important because the ration used to express the concentration of  any  solution.

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The results of the lab activity showed that the larger the mass of the sun, the more likely at least one planet will fall into the habitable zone.

The mass of the sun affects the orbits of planets in a solar system. When the mass of the sun is larger, the gravitational force between the sun and the planets is stronger, causing the planets to move at a slower pace around the sun.

Conversely, when the mass of the sun is smaller, the gravitational force is weaker, causing the planets to move at a faster pace.

Additionally, when Earth is closer to the sun, the gravitational force is stronger, causing its orbit to become faster, while a farther distance from the sun results in a slower orbit.

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Answer: 6.11 L

Explanation:

STP= 1atm, 273.15K

Molar mass of CO2=44.01g/mol so n= (12.0/44.01)

PV=nRT

V=(nRT)/P

V=((12.0/44.01)(0.0821)(273.15))/1

V=6.11L

Answer:

The Correct answer is option 3

Step by Syep Explanation:

4Fe+3O2---->rust

formula for rust----->Fe2O3

4Fe+3O2---->Fe2O3

Balancing the Chemical Equation

both the reactant and product side

we have that;

4Fe+3O2------->2FeO3

the equation is Chemically Balanced

therefore 4Fe+3O2------->2×rust

Answer:

Explanation:

Given that,

Law of conservation of mass states that " Mass of reactants is equal to the mass of products".

Also we know that In a balanced equation the total number of atoms in the reactants equals the total number of atoms in the product.

We are given with 4Fe + 3O₂ i.e the reactant.

First Let's calculate the number of atoms in the reactant.

Now, Let's find the product .

Also, We can see 2Fe₂O₃ (Product)

4Fe + 3O₂ → 2Fe₂O₃

Number of atoms in the reactants = the total number of atoms in the product.

Therefore, 2Fe₂O₃ (Option 3) will the required answer .

Using the Ideal Gas Law, the pressure expected for 30 mol of  [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.

Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)

Solve for P (pressure):

P = nRT / V

P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L

P = 1210.07 atm

Van der Waals equation:

(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol

Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)

P = 1179.71 atm

Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

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Answer:

90g

Explanation:

Ans. 90 gram

we know that,

n = wt/m.wt

where, n=  moles

wt.= weight

m.wt = molecular weight

putting values we get

5 = wt./18 ( molecular weight of water is 18

wt.= 90

hence  ans.= 90 gram

The standard cell potential is found as +1.95 V and is a  spontaneous  reaction.

The standard cell potential (E°cell) of an electrochemical cell is given by the difference between the standard reduction potentials of the two half-cells involved.

E°cell = E°reduction (cathode) - E°reduction (anode)

The half-reactions given are:

Pb4+ + 2e− → Pb2+ (reduction)

Co3+ + e− → Co2+ (reduction)

The standard reduction potentials for these half-reactions are:

E°reduction(Pb4+/Pb2+) = -0.13 V

E°reduction(Co3+/Co2+) = +1.82 V

We then calculate as:

E°cell = E°reduction (Co3+/Co2+) - E°reduction (Pb4+/Pb2+)

E°cell = (+1.82 V) - (-0.13 V)

E°cell = +1.95 V

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