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soldi70 [24.7K]
3 years ago
12

Determine the enthalpy change when 23.5 g of carbon is reacted with oxygen according to the reaction C(s) + O2(g) CO2(g) delta H

= –394 kJ/mol
Chemistry
1 answer:
xeze [42]3 years ago
8 0
We are provided with the amount of energy released when one mole of carbon reacts. We mus first convert the given mass of carbon to moles and then compute the energy released for the given amount.

Moles = mass / atomic mass
Moles = 23.5 / 12
Moles = 1.96 moles


One mole releases 394 kJ/mol
1.96 moles will release:
394*1.96
= 772.24

The enthalpy change of the reaction will be -772.24 kJ
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What volume would a 0. 250 mole sample of h2 gas occupy if it had a which a a pressure of 1. 70 atm and a temperature of 35 C?
frosja888 [35]

The volume of  0. 250 mole sample of H_{2} gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is  3.71 L.

Calculation,

According to ideal gas equation which is known as ideal gas law,

PV =n RT

  • P is the pressure of the hydrogen gas  = 1.7 atm
  • Vis the volume of the hydrogen gas = ?
  • n is the number of the hydrogen gas = 0.25 mole
  • R is the universal gas constant = 0.082 atm L/mole K
  • T is the temperature of the sample = 35°C = 35 + 273 = 308 K

By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,

1.7 atm×V = 0.25 mole ×0.082 × 208 K

V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm

V = 3.71 L

So, volume of the sample of the hydrogen gas occupy is  3.71 L.

learn more about ideal gas equation

brainly.com/question/4147359

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4 0
1 year ago
Ammonia, NH3; ammonium nitrate, NH4NO3; and ammonium hydrogen phosphate, (NH4)2HPO4, are all common fertilizers. Rank the compou
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Answer:

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

Explanation:

Mass percentage -

Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.

i.e.

mass % A = mass of A / mass of solution * 100

Given,  

mass of nitrogen = 14 g/mol

mass of hydrogen = 1 g/mol

mass of oxygen = 16 g/mol

mass of phosphorus = 31 g/mol

1. Ammonia, NH₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 1 * 14 g/mol  + 3 * 1 g/mol  

mass of solution = 17 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 17 g/mol * 100

mass % N = 82.35 % .

2.   ammonium nitrate, NH₄NO₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 4 * 1 g/mol  + 3 * 16 g/mol

mass of solution = 80 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 80 g/mol * 100

mass % N = 17.50 % .

3.   ammonium hydrogen phosphate, (NH₄)₂HPO₄

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 9 * 1 g/mol  + 4 * 16 g/mol + 1 * 31 g/mol

mass of solution = 132 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 132 g/mol * 100

mass % N = 10.60 % .

Hence , the correct order is -

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

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What is the relationship between the number of each atom used to form compound and oxidation number?
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