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lord [1]
3 years ago
7

What maximum amount of ammonia in kilograms can be synthesized from 5.22 kg of h2 and 31.5 kg of n2?

SAT
1 answer:
Fiesta28 [93]3 years ago
4 0

Consider the unbalanced reaction,

H₂ + N₂   ===>   NH₃

Balance it to get

3 H₂ + N₂   ===>   2 NH₃

This tells you that for every 3 moles of H₂ and 1 mole of N₂ among the reactants, 2 moles of NH₃ are produced.

Look up the molar masses of each component element:

• H : 1.008 g/mol     ===>   H₂ : 2.016 g/mol

• N : 14.007 g/mol     ===>   N₂ : 28.014 g/mol

• NH₃ : 17.031 g/mol

Convert the given mass to moles for each reactant:

• H₂ :

(5.22 kg) × (1000 g/kg) × (1/2.016 mol/g) ≈ 2589.29 mol

• N₂ :

(31.5 kg) × (1000 g/kg) × (1/28.014 mol/g) ≈ 1124.44 mol

Now,

(2589.29 mol H₂) : (1124.44 mol N₂) ≈ (2.30 mol H₂) : (1 mol N₂)

so that the amount of N₂ consumed is

(2589.29 mol H₂) × (1 mol N₂) / (3 mol H₂) ≈ 863.095 mol N₂

leaving an excess of about 261.343 mol N₂, and producing

(2589.29 mol H₂) × (2 mol NH₃) / (3 mol H₂) ≈ 1726.19 mol NH₃

Convert this to a mass :

(1726.19 mol) × (17.031 g/mol) × (1/1000 kg/g) ≈ 29.3987 kg

So, up to 29.4 kg of NH₃ can be synthesized.

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The store that has a greater variety of numbers of wristbands sold is the store whose box plot has a greater IQR value.

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  • See the diagram attached below to understand how to get the Q3 and Q1 of the data distribution.

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It should be noted that the Great Compromise combined elements of Virginia and New Jersey plan.

According to this question, we are to discuss the regions that serves as the combined elements of Great Compromise.

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Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca
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The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t} (1)

Where:

  • T_{s} - Final temperature, in degrees Celsius.
  • T_{o} - Initial temperature, in degrees Celsius.
  • t - Time, in minutes.
  • k - Cooling rate, in \frac{1}{min}.
  • T(t) - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when T(t) = T_{s}, that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that T_{o} = 80\,^{\circ}C, t = 15\,min, T_{s} = 50\,^{\circ}C and T(15) = 60\,^{\circ}C, then the cooling rate of the substance is:

60 = 50 + (80 - 50)\cdot e^{-15\cdot k}

\frac{60-50}{80-50}= e^{-15\cdot k}

\frac{1}{3} = e^{-15\cdot k}

k = -\frac{1}{15}\cdot \ln \frac{1}{3}

k \approx 0.0732

The cooling rate of the substance is approximately 0.0732.

To learn more on Newton's law of cooling, we kindly invite to check this verified question: brainly.com/question/13748261

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