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Bas_tet [7]
3 years ago
6

A 2.10 L vessel contains 4.65 mol of nitrous oxide (N2O) at 3.82 atm. What will be the pressure of this gas in a container that’

s half the size?
Chemistry
1 answer:
sveticcg [70]3 years ago
4 0

Answer:

The answer to your question is 7.64 atm

Explanation:

Data

Volume 1 = V1 = 2.1 l

moles = 4.65

Pressure 1 = P1 = 3.82 atm

Volume 2 = Volume 1/2

Pressure 2 = ?

Process

1.- Calculate the new volume

Volume 2 = 2.10/2

Volume 2 = 1.05 l

2.- Use Boyle's law to find the Pressure

             P1V1 = P2V2

-Solve for P2

              P2 = P1V1 / V2

-Substitution

              P2 = (3.82)(2.1) / 1.05

-Simplification

              P2 = 8.022/1.05

-Result

              P2 = 7.64 atm

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7 0
3 years ago
Describe the particles in each state of matter <br><br> Solid:<br> Liquid: <br> Gas:
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3 years ago
After being thoroughly stirred at 10.°C, which mixture is heterogenous?(1) 25.0 g of KCl and 100. g of H2O
zheka24 [161]
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5 0
3 years ago
Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
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