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<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435
Answer:
The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.
Explanation:
2S (s) + 3O₂ (g) → 2SO₃ (g)
64g + 96g → 160 g
32g + 48g → 80 g
x + y → 100 g
1 mol SO₃ ___ 80g
n _______ 100g
n = 1.25 mol SO₃
1 mol S ___ 32 g
1,25 mol S __ 40 g
1 mol O₂ ___ 32 g
1,875 mol O₂ ___ 60 g
Let's start with the amount given in percent. Let our basis be 100 grams of compound. So, that means that in this amount, 57.1 g is oxygen and 100-57.1=42.9 g is carbon. Since there is 1:1 atom ratio, it also means that moles oxygen = moles carbon.
Moles = Mass/Relative Mass
Let x be the relative mass of oxygen
57.1/16 = 42.9/x
Solving for x,
<em>x = 12.02 amu</em>