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3 years ago

During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.

Chemistry

1 answer:

Brrunno [24]3 years ago

3 0

Answer: 19.4 mL Ba(OH)2

Explanation:

H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)

At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.

0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2

Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.

0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl

HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2

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<h3>Answer:</h3>

1000 g CCl₄

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

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

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})$
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<u>Step 4: Check</u>

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1003.77 g CCl₄ ≈ 1000 g CCl₄

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3 years ago

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Explanation:

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2 years ago

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3 years ago

How many liters of a 6.0M solution Fe(SCN)2 are needed to prepare 2.0L of a 0.75M solution?

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Answer:

0.25L

Explanation:

Using the dilution formula

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3 years ago

Read 2 more answers

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar. part a part complete how many milligrams of sugar does the m

kolezko [41]

Answer:
A 12.630 bar of chocolate contains 8315 milligrams of sugar

Explanation:
From the basics of conversion, we can find that:
1 gram is equal to 1000 mg
Using cross multiplication, we can find how many milligrams are present in 8.315 grams as follows:
1 gram .............> 1000 mg
8.315 grams ....> ?? mg
amount in milligrams = (8.315*1000) / 1 = 8315 milligrams

Hope this helps :)

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3 years ago