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Karo-lina-s [1.5K]
3 years ago
14

The molar heat capacity of mercury is 28.1j/mol•k what is the specific heat capacity of metal in j/g•k

Chemistry
1 answer:
rusak2 [61]3 years ago
5 0
M_{Hg}=201\frac{g}{mol}\\\\
c=28.1\frac{J}{mol*K}=28,1\frac{J}{201\frac{g}{mol}*mol*K}\approx0,1398\frac{J}{gK}
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How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
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The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

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