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4 years ago

How many chlorine molecules are in 7.02 l of chlorine gas at stp?

1 answer:

5 0

The answer is: [B]: 1.89 * 10²³ .
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PV = nRT ; is the equation for "STP" conditions; that is, the "ideal gas equation" .
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i.e. when Pressure, "P" = 1.00 atm;
Temperature, "T" = 273 K;
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R = the ideal gas constant = ((0.08206 L-atm/K-mol)
n = number of moles;
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So, we plug in our known values:
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(1.00atm) (7.02L) = ("n" mol) (0.08206 L-atm/K-mol) (273K); 
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→ 7.02 L·atm = (? mol) (22.4 L·atm/mol) .

(Note that the Molar Volume of a gas at STP is a constant using Avogadro's value of 22.4 L / mol. 1 mol of any ideal gas at STP occupies 22.4 L. An ideal gas takes the shape of its container)..
_______________________________________________________
 → Divide EACH side of the equation by "(22.4 L·atm/mol)" ;
________________________________________________
→ 7.02 L·atm / = ("n" mol) (22.4 L·atm/mol) ;
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→ 7.02 L·atm / (22.4 L·atm/mol) =
[("n" mol) (22.4 L·atm/mol)]/(22.4 L·atm/mol);
______________________________________________________
to get:
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→ n = 0.313 mol ;
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Note: 1 mole = 1 mol = 6.022 * 10²³ molecules.
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→ So, 0.313 mol Cl₂ (g) molecules * [(6.022 * 10²³ molecules) / (1 mol)] =
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→ [(0.313) * (6.022 *10²³) ] molecules of Cl₂ (g) ;
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→ = 1.88 * 10²³ molecules of Cl₂ g ;
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→ which most closely corresponds with answer choice:
___________________________________________________________
[B]: 1.89 * 10²³ .
___________________________________________________________

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Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

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Step 8: Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

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