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Nata [24]
2 years ago
9

You are given the reaction Cu + HNO3 → Cu(NO3)2 + NO + H2O. Half-reactions: First: 3 upper C u right arrow 3 upper C u superscri

pt 2 plus, plus 6 e superscript minus. Second: 2 upper N superscript 5 plus, plus 6 e superscript minus right arrow 2 upper N superscript 2 plus. Complete the final balanced equation based on the half-reactions. Cu + HNO3 Right arrow. Cu(NO3)2 + NO + H2O
Chemistry
2 answers:
Alona [7]2 years ago
8 0

Answer:

Explanation:

You are given the reaction Cu + HNO3 → Cu(NO3)2 + NO + H2O.

Half-reactions:

First: 3 upper C u right arrow 3 upper C u superscript 2 plus, plus 6 e superscript minus. Second: 2 upper N superscript 5 plus, plus 6 e superscript minus right arrow 2 upper N superscript 2 plus.

Complete the final balanced equation based on the half-reactions.

⇒ 3Cu +  

⇒ 8HNO3

⇒ 3Cu(NO3)2 +  

⇒ 2NO +  

⇒ 4H2O

nexus9112 [7]2 years ago
4 0

Answer: It is the first one. First: 3 upper C u right arrow 3 Upper C u superscript 2 plus, plus 6 e superscript minus. Second: 2 upper N superscript 5 plus, plus 6 e superscript minus right arrow 2 upper N superscript 2 plus.

Explanation:

Correct

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Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
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Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

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Molarity = \frac{n}{V(L)}

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V = 2.13084 L

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Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

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