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2 years ago

You have 1 mole of a gas at STP. If you apply the ideal gas law what is the approximate volume of the gas?

Chemistry

1 answer:

goldfiish [28.3K]2 years ago

7 0

Answer:

A) 22.4L

Explanation:

we know, ideal gas law states

PV=nRT

V=nRT/P

At STP,

T= 273.15K P=1atm R=0.082L.atm/mol/K n=1 mole

V=(1*0.082*273.15)/ 1

V=22.4L

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Which elements on the periodic table fulfill the octet rule

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Answer:

Noble Gases

Explanation:

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3 years ago

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

412.072 g Cl₂ ≈ 412 g Cl₂

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2 years ago

What volume would 3.01•1023 molecules of oxygen gas occupy at STP?

Sliva [168]

First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2

Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2

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