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MissTica
3 years ago
5

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

ion to determine the mass of NaCl that reacted with F2 at 290. K and 1.5 atm.
F2 + 2NaCl → Cl2 + 2NaF

Part 2. Explain how you would determine the mass of sodium chloride that can react with the same volume of fluorine gas at STP.
Chemistry
1 answer:
Alika [10]3 years ago
8 0

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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Answer:

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Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

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v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

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Km will still be much smaller then [S], so

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v = 0.443 μmol/min

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Why must each atom of an element always have the same number of protons?
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8 0
3 years ago
If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?
Ket [755]
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If one mole of Fe reacts with two moles of sodium 
Then 4 moles of Fe produces 8 moles of sodium.
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stich3 [128]

Answer:

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4 0
3 years ago
6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm
Stels [109]

<u>Answer:</u> The molar mass of the gas is 35.87 g/mol.

<u>Explanation:</u>

To calculate the mass of gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of gas = 945 mmHg

V = Volume of the gas = 0.35 L

m = Mass of gas = 0.527 g

M = Molar mass of gas = ? g/mo

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of gas =  88^oC=[88+273]=361K

Putting values in above equation, we get:

945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol

Hence, the molar mass of the gas is 35.87 g/mol.

5 0
3 years ago
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