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GrogVix [38]
2 years ago
9

The gases listed are at standard temperature and pressure: · 1 mol of CO2 gas · 1 mol of N2 gas · 2 mol of O2 gas Which statemen

t is correct? The volume of 1 mol of CO2 is greater than that of 2 mol of O2 . The volume of 1 mol of CO2 is greater than that of 1 mol of N2 . The volumes of 2 mol of O2 and 1 mol of N2 gases are the same. **The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same.**
Chemistry
2 answers:
andrezito [222]2 years ago
8 0
The right answer for the question that is being asked and shown above is that: "The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same." The <span>statement that is correct as far as the </span>gases listed are at standard temperature and pressure is concerned is that t<span>he volumes of 1 mol of CO2 and 1 mol of N2 gases are the same</span>
Elena L [17]2 years ago
6 0

Answer : The volume of 1 mole of CO_{2} and N_{2} gases are the same.

Solution: Given,

Moles of CO_{2} = 1 mole

Moles of N_{2} = 1 mole

Moles of O_{2} = 2 mole

Formula used for ideal gas is :

P V = n R T

According to the question, the gases are at standard temperature and pressure. So, the volume of gases only depends on the number of moles. This means that the higher the number of moles, higher will be the volume of gas.

The moles of O_{2} are more than the moles of CO_{2} and N_{2}. So, the volume of O_{2} will be more.

And the moles of CO_{2} and N_{2} are equal. Therefore, their volumes are also equal.

Therefore, the best option is the volume of 1 mole of CO_{2} and N_{2} gases are the same.

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Solving this chemistry is a little bit hard because the question didn't give some important detailed. 
So first, there are a couple problems with your question. 
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be: 
Kc = [C] / [A] [B]^2 
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.
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What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?
Step2247 [10]

Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

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3 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0
3 years ago
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