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3 years ago

Moles of Zn(NO3)2 in 131.50g of this substance.

Chemistry

2 answers:

ch4aika [34]3 years ago

5 0

Answer:

The quantity of moles in 131.50g of Zn(NO3)2 is 0.6943

Explanation:

Using the equation:

Moles = mass/ molecular mass

Moles Zn(NO3)2 = mass Zn(NO3)2 / molecular mass Zn(NO3)2

mass Zn(NO3)2 = 131.50g

molecular mass Zn(NO3)2 is calculated by looking at the atomic mass of each element in the periodic table, then multiplying by the number on the compound's formula:

Zn= 65.38 g/mol

N= 14 g/mol

O= 16 g/mol

molecular mass Zn(NO3)2 = Zn + 2*N + 2*3*O =65.38 + (2*14) + (2*3*16)

molecular mass Zn(NO3)2 = 189,38 g/mol

Moles Zn(NO3)2 = 131.50g / 189,38 g/mol

Moles Zn(NO3)2 = 0.6943

Degger [83]3 years ago

3 0

number of moles= mass / molecular mass
mass=131.50g
molecular mass of Zn(NO3)2 =189.39 g/mol
number of moles = 131.50 g/189.39 g/mol =0.69433 mol

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For B: The $K_c$ for the given reaction is 1642.

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$K_c$ = equilibrium constant in terms of concentration = ?

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Putting values in above equation, we get:

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