3. Given the following equation:
1 answer:
Answer:
4.767 grams of KCl are produced from 2.50 g of K and excess Cl2
Explanation:
The balanced equation is
2 K+ Cl2 --->2 KCI
Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K
Mass of one atom/mole of potassium is 39.098 grams
Number of moles is 2.5 grams =
So, 2 moles of K produces 2 moles of KCL
0.064 moles of K will produces 0.064 moles of KCl
Mass of one molecule of KCl is 74.5513 g/mol
Mass of 0.064 moles of KCl is 4.767 grams
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The mass of steam required to raise the temperature of water is 3.5 g.
The given parameters;
- <em>mass of the benzene, = 47.6</em>
- <em>initial temperature of the benzene, = 5.5 ⁰C</em>
- <em>final temperature of the benzene = 30 ⁰C</em>
The molar mass of Benzene = 78.11 g/mol
The molar mass of water = 18 g/mol
The number of moles of the Benzene is calculated as follows;
The mass of steam required is calculated as follows;
<em>heat lost by steam = heat absorbed by benzene</em>
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Thus, the mass of steam required to raise the temperature of water is 3.5 g.
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Answer: 5.85kJ/Kmol.
Explanation:
The balanced equilibrium reaction is
The expression for equilibrium reaction will be,
Now put all the given values in this expression, we get the concentration of methane.
Relation of standard change in Gibbs free energy and equilibrium constant is given by:
where,
R = universal gas constant = 8.314 J/K/mole
T = temperature =
= equilibrium constant = 10.6
Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.