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photoshop1234 [79]
3 years ago
12

copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

Chemistry
1 answer:
a_sh-v [17]3 years ago
8 0

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:

2KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

Learn more about balancing chemical equations: brainly.com/question/8062886

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Solution:- This problem is based on Clausius clapeyron equation--

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Let's plug in the values in the equation and do the calculations.

ln(\frac{400}{P_2})=(\frac{39300}{8.314})(\frac{1}{307.9}-\frac{1}{336.5})

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On taking anti ln to both sides...

\frac{400}{P_2} = e^1^.^3^0

\frac{400}{P_2} = 3.67

P_2 = 400/3.67

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