What mass of ZnCO3 contains 3.11×1022 O atoms

Chemistry

2 answers:

eduard [2.7K]3 months ago

5 0

From the molecular formula of the compound, given as $ZnCO_{3}$ , it’s evident that 1 mole of $ZnCO_{3}$ consists of 3 moles of oxygen.

In one mole, there are $6.023\times 10^{23}$ atoms, thus, having 3 moles of oxygen corresponds to $18.069\times 10^{23}$ atoms of oxygen.

This means that 1 mole of $ZnCO_{3}$ contains $18.069\times 10^{23}$ oxygen atoms.

1 atom of oxygen is equivalent to $3\times 6.023\times 10^{23}=5.53\times 10^{-25}$ moles.

From the number of oxygen atoms provided, which is $3.11\times 10^{22}$ , it will be contained in $3.11\times 10^{22}\times 5.53\times 10^{-25} mol.=0.01721 mol$ moles of $ZnCO_{3}$ .

The molar mass of $ZnCO_{3}$ is 125.388 g/mol, enabling us to compute the mass as follows:

$n=\frac{m}{M}$

By rearranging the equation above,

$m=n\times M$

Substituting the values gives us,

$m=(0.01721 mol)\times (125.388 g/mol)=2.16 g$

Thus, the mass of $ZnCO_{3}$ amounts to 2.16 g.

KiRa [2.9K]3 months ago

3 0

Please ask if you have any inquiries.

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Some oxygen gas takes up a volume of 2.00 liters at 0.99 atm and 273 K. Its volume doubles and its temperature decreases to 137

KiRa [2933]

The response to your inquiry is: option B. 0.25 atm

Explanation:

To solve this issue, the combined gas law must be applied:

P₁V₁ = P₂V₂ / T₁T₂

The data is as follows: P1 = 0.99 atm, V1 = 2 L, T1 = 273 K, P2 =?, V2 = 4 L, T2 = 137 K.

By isolating P2 in the equation, you find

P2 = P1V1T2 / T1V2. Substituting in the numbers gives: P2 = (2 x 0.99 x 137)/(273 x 4). The resulting P2 equates to approximately 0.25 atm.

5 0

3 months ago

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