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9966 [12]
4 years ago
10

What mass of zinc oxide would be produced by the thermal decomposition of 375 grams of zinc carbonate?

Chemistry
1 answer:
Llana [10]4 years ago
6 0
Zinc carbonate has the chemical formula : ZnCO3 and the equation for thermal decomposition is:
ZnCO3 .................> ZnO + CO2
1 mole of ZnCO3 produces one mole of ZnO

From the periodic table:
molar mass of zinc = 65.38 gm
molar mass of carbon = 12 gm
molar mass of oxygen = 16 gm

molar mass of ZnCO3 = 65.38 + 12 + 3(16) = 125.38 gm
molar mass of ZnO = 65.38 + 16 = 81.38 gm

125.38 gm of ZnCO3 produces 81.38 gm of ZnO, therefore:
mass of ZnO in 375 gm = (375 x 81.38) / 125.38 = 243.4 gm

Based on the above calculations, the correct answer is 243 grams


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What volume does 4.24g of argon gas occupy at 58.2 degrees celcuis and 1528torr
Radda [10]

The volume occupied by the argon gas is, 1.44 L

Explanation :

To calculate the volume of argon gas we are using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of argon gas = 1528 torr = 2.010 atm

Conversion used : (1 atm = 760 torr)

V = volume of argon gas = ?

T = temperature of argon gas = 58.2^oC=273+58.2=331.2K

R = gas constant = 0.0821 L.atm/mole.K

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Now put all the given values in the ideal gas equation, we get:

(2.010atm)\times V=\frac{4.24g}{39.95g/mole}\times (0.0821L.atm/mole.K)\times (331.2K)

V=\frac{4.24g}{39.95g/mole}\times \frac{(0.0821L.atm/mole.K)\times (331.2K)}{2.010atm}

V=1.44L

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7 0
3 years ago
the pressure of 1 mol of gas is decreased to 0.05 atm 273.k. what happens to the molar volume of the gas under these conditions?
svp [43]

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<h3>Further explanation</h3>

Given

n = 1 mol

P = 0.05 atm

T = 273 K

Required

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PV=nRT

V = nRT/P

Input the value :

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