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2 years ago

How many mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound?

Chemistry

1 answer:

saul85 [17]2 years ago

6 0

2.87mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a substance can be calculated by dividing the number of moles of the substance by its molarity. That is;

volume = no. of moles ÷ molarity

According to this question, 2.5M K2SO4 are required to obtain 1.25 grams of the compound. The volume can be calculated as follows:

Molar mass of K2SO4 = 174.26 g/mol

moles of K2SO4 = 1.25g ÷ 174.26g/mol

moles of K2SO4 = 0.00717moles

Volume = 0.00717moles ÷ 2.5

Volume = 0.00287 L = 2.87mL.

Therefore, 2.87mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound.

Learn more about volume at: brainly.com/question/1578538

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Explanation:

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A electrochemical reaction is said to be spontaneous, if $E^{0} cell is positive.$

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

The cell representation of above reaction is given by;
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Answer 2:
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
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The cell representation of above reaction is given by;
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Hence, $E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}$
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Answer 3:
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
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The cell representation of above reaction is given by;
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Hence, $E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}$
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