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slega [8]
3 years ago
13

Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. cons

ider the other product and its phase, and then write the balanced molecular equation for this precipitation reaction
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
4 0

Lithium sulfate is Li2SO4

Strontium chlorate is Sr(ClO3)2

Strontium sulfate is SrSO4

 

So the complete balanced chemical reaction for this is:

 

Li2SO4 (aq) +  Sr(ClO3)2 (aq) -->  SrSO4 (s)  +  2 LiClO3 (aq)

 

 

This is a type of double replacement reaction since there is an exchange of ions.

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Which statement best describes the pH of pure water? It is neutral because the concentration of hydronium ions equals that of hy
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The pH of pure water has been best described as neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

pH has been described as the measurement of hydrogen ions in a solution. The pH has been measured on a scale of 1-14. pH 7 has been the neutral pH.

The higher hydronium ion concentration tends to move the pH from 7 towards 1. The higher hydroxide ion concentration tends to move the pH above 7.

The neutral pH has been neither acidic nor basic with the equal constituents of hydronium and hydroxide ion in the solution.

Thus, the pH of pure water has been 7. It has neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

For more information about the pH of the solution, refer to the link:

brainly.com/question/4975103

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B 144.0 s is the best answer of this question

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the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
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<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

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\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

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3 years ago
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Bezzdna [24]
Huh. Is this supposed to be biology?
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2 years ago
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