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3 years ago

How many atoms are present in a sample of Calcium (Ca) weighing 127.38g?

Chemistry

1 answer:

kari74 [83]3 years ago

3 0

Answer:

1.9139 × 10²⁴ atoms Ca

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

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

Explanation:

<u>Step 1: Define</u>

127.38 g Ca

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

Avogadro's Number

Molar Mass of Ca - 40.08 g/mol

<u>Step 3: Convert</u>

<u /> $127.38 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca} )(\frac{6.022 \cdot 10^{23} atoms \ Ca}{1 \ mol \ Ca} )$ = 1.91388 × 10²⁴ atoms Ca

<u>Step 4: Check</u>

<em>We are given 5 sig figs. Follow sig fig rules and round.</em>

1.91388 × 10²⁴ atoms Ca ≈ 1.9139 × 10²⁴ atoms Ca

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A sample of unknown hydrate, AC-XH20, has a mass of 1.000 g before heating and a

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Answer:

The water of crystallization for the formula of the unknown hydrate is 2.

Explanation:

Mass of an unknown hydrate = 1.000 g

Molar mass of hydrate = 195.5 g/mol

MOles of unknown hydrate = $\frac{1.000 g}{195.5 g/mol}=0.00512 mol$

Mass of hydrate after heating = 0.781 g

Mass of water lost due to heating = x

$1.000 g = 0.781 g + x\\x = 1.000 g - 0.781 g = 0.219 g$

Moles of water lost = $\frac{0.219 g}{18.01528 g/mol}=0.0122 mol$

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1 Mole of hydrate has x moles of water and 0.00512 moles of hydrate has 0.0122 moles of water then we can write:

$\frac{1 }{x}=\frac{0.00512 }{0.0122 }\\x\approx 2$

The water of crystallization for the formula of the unknown hydrate is 2.

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3 years ago

PLEASE HELP FAST! When 23.3 g of O2 reacts with 18.3 g C10H8, what is the limiting reactant. The equation is C10H8 + 12O2 -->

Murljashka [212]

O₂ is the limiting reagent.

<h2>What is a limiting reactant?</h2>

A limiting reactant is described as the one that will be consumed first in a chemical reaction.

<h3>Calculation</h3>

C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O

Given, Mass of O₂ = 23.3 g

Mass of C₁₀H₈ = 18.3 g

Molar mass of O₂ = 32 g

No. of moles of O₂ = 32/23.3 = 1.4 moles

Molar mass of C₁₀H₈ = 128 g

No. of moles of C₁₀H₈ = 128/18.3 = 6.9 = 7 moles

According to the equation,

1 mole of C₁₀H₈ yields 10 moles of CO₂. So, 7 moles of C₁₀H₈, give 70 moles of CO₂.

12 moles of O₂ yields 10 moles of CO₂. So, 1.4 moles of O₂ give 1.16 moles of CO₂.

We do know, however, that based on our balanced equation, if we were to totally react all 1.4 moles of O2, we could only produce a maximum of 1.16 moles of CO2. Even while there is enough C10H8 to generate more, the amount of O2 we have is a limiting factor because it will be consumed first.

So, our limiting reactant is O₂.

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Answer:

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Given data:

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