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Thepotemich [5.8K]
2 years ago
14

How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical curre

nt of 12.0 a? a. 1.19 ´ 103 b. 27.0 c. 2.90 ´ 105 d. 9.00 e. 3.57 ´ 103?
Chemistry
1 answer:
Brilliant_brown [7]2 years ago
8 0
The correct answer is e. 3.57×10³
Al³+(aq) + 3e→AL(s)
4.00g of AL=4g/26.98 g/mol= 0.1483 mol
t=znF/1 where t is time in seceonds.
Z= valency number of ions of the substance or electrons which are transferred per ion
F= Faraday's constant
I = electric current in'A'CA  C/s
t=(3×0.1483 mol ×96485 C/mol) /12(C15)
t=3577 second = 3.5 ×10³s
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They differ in their molecular structures and properties.
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What is the boiling point of cyclohexane at 620 mmhg? can you show calculations?
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 Boiling point of cyclohexane at 620 mm hg?  

Standard atmospheric pressure is 760 mm Hg. Boiling point at 760 mm= 80.74˚C  

A liquid boils when its vapor pressure is equal to the atmospheric pressure. The vapor pressure of a liquid is proportional the absolute temperature of the liquid.  

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The boiling point of cyclohexane at 620 mm Hg is less than 80.74˚C



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What is the mass of 1.6x1020 molecules of carbon dioxide?
valentina_108 [34]

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1.2*10^{-2} g

Explanation:

1 mol - 6*10^{23} particles

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8 0
2 years ago
Consider the following intermediate chemical equations.
vfiekz [6]

Answer:

-250.3kJ

Explanation:

Based in the reactions and using -<em>Hess's law-</em>:

(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ

(2) 4PCl₅(g) → P₄(s) + 10Cl₂  ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ

Dividing this reaction in 4:

PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>

8 0
2 years ago
All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou
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Answer:

m_{Mg}=30.8mgMg

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg

Best regards.

3 0
2 years ago
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