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3 years ago

A sample of 1.00 moles of oxygen at 50°C and 98.6 kPa, occupies what volume?

Chemistry

1 answer:

enyata [817]3 years ago

5 0

Answer:

27.22 dm³

Explanation:

Given parameters:

number of moles = 1 mole

temperature= 50°C, in K gives 50+ 273 = 323K

Pressure= 98.6kpa in ATM, gives 0.973 ATM

Solution:

Since the unknown is the volume of gas, applying the ideal gas law will be appropriate in solving this problem.

The ideal gas law is mathematically expressed as,

Pv=nRT

where P is the pressure of the gas

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

Input the parameters and solve for V,

0.973 x V = 1 x 0.082 x 323

V= 27.22 dm³

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How many grams of Fe will be produced if 705.0g of H20 are produced?

fredd [130]

<h3>Answer:</h3>

2265 g Fe₃O₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

<u>Step 2: Identify Conversions</u>

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

<u>Step 3: Convert</u>

Set up stoich: $\displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})$
Multiply/Divide/Cancel units: $\displaystyle 2264.74 \ g \ Fe_3O_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

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