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klio [65]
2 years ago
5

If the mass ratio of nitrogen to hydrogen in ammonia is 4.7:1. if a sample of ammonia contains 10.0 gram of h, how many grams of

n does it contain?
Chemistry
1 answer:
Tju [1.3M]2 years ago
6 0

The amount, in grams, of N that the sample will contain will be 2.1 grams.

<h3>Stoichiometric mass ratio</h3>

According to the problem. the mass ratio of H and N in ammonia is 4.7:1.

In other words, every 4.7 grams of H in ammonia must have 1 gram of N.

Now, in a particular ammonia sample, 10 grams of H is present.

The amount of N present can be calculated from the standard mass ratio.

4.1 grams H = 1 gram N

10 grams H = x

4.1x = 1 x 10

        x = 10/4.1

             x = 2.1 grams

Thus, the mass of nitrogen in the ammonia sample will be 2.1 grams.

More on mass ratios can be found here: brainly.com/question/14577772

#SPJ1

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5 0
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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
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Answer : The rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-2}M^{-2}s^{-1}

Thus, the value of the rate constant 'k' for this reaction is 7.5\times 10^{-2}M^{-2}s^{-1}

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1

\text{Rate}=4.25\times 10^{-2}Ms^{-1}

Therefore, the rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

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