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sdas [7]
3 years ago
12

1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by

mass. Further analysis determines the molar mass to be 30.04g/mol. Determine the molecular formula of the unknown.
*please show step by step process!!!*
Chemistry
1 answer:
Nitella [24]3 years ago
6 0

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

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2 years ago
Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium
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Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

k_{c}=3.07\times10^{-4}

Temperature = 24°C

We know that,

The balance equation is

2NOBr\Rightarrow 2NO+Br_{2}

Initial concentration is,

0.878\Rightarrow 0+0

Concentration is,

-2x\Rightarrow 2x+x

Equilibrium concentration

0.878-2x\Rightarrow 2x+x

We need to calculate the value of x

Using formula of concentration

k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}

Put the value into the formula

3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}

2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}

2x^2=0.0002367+0.001228x^2-0.0010536x

2x^2-0.001228x^2+0.0010536x-0.0002367=0

1.998772x^2+0.0010536x-0.0002367=0

x=0, 0.01062

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

concentration\ of\ NO=2x

Put the value of x

concentration\ of\ NO=2\times0.01062

concentration\ of\ NO=0.02124

Hence, The equilibrium concentration of NO is 0.02124 M.

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