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3 years ago

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The theoretical yield of ammonia in an industrial synthesis was 550 kg, but only 480 kg was obtained. What was the percentage yi

eld of the reaction?

1 answer:

bezimeni [28]3 years ago

8 0

Theoretical yield 550 kg - 100%

Actual yield 480 kg - x%

x= 480*100/550 ≈ 87.3 %

Percentage yield ≈ 87.3 %.

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In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

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Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

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2 years ago

If moving objects have kinetic energy, do moving atoms kinetic energy?

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2 years ago

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

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3 years ago