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MAVERICK [17]
3 years ago
6

Use this balanced equation to help you solve:

Chemistry
1 answer:
valentina_108 [34]3 years ago
5 0

1. 842g of NaOH will form 547.3 g of Al(OH)₃

2. The yield is 93.55%

<u>Explanation:</u>

3NaOH + Al → Al(OH)₃ + 3Na

1.

Molar mass of NaOH = 40 g/mol

Molar mass of Al = 27 g/mol

Molar mass of Al(OH)₃ = 78 g/mol

According to the balanced equation:

3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃

The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1

3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃

120 g of NaOH + 27g of Al → 78 g of Al(OH)₃

120g of NaOH form 78g of Al(OH)₃

1g of NaOH will form \frac{78}{120} g of Al(OH)₃

842g of NaOH will form \frac{78}{120} X 842 g of Al(OH)₃

                                   = 547.3 g of Al(OH)₃

Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃

2. Only 512 g of Al(OH)₃ is formed

Yield % = ?

Yield = \frac{512}{547.3} X 100\\\\Yield = 93.55

Therefore, the yield is 93.55%

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bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

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Given ,  

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P₁ = 2575 / 760 atm = 3.39 atm

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T₁ = 353 K

T₂ = 253 K

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}

\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}

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45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0
3 years ago
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Answer:

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Explanation:

Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.

The amount of NaCl required is calculated as follows:

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<u>Given:</u>

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<u>To determine:</u>

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T1 = 20 + 273 = 293 K

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3 0
3 years ago
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Answer:

Explanation:

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