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fenix001
2 months ago
14

Why couldn't you substitute 3m H2SO4 for concentrated HNO3 when oxidizing copper?

Chemistry
2 answers:
lorasvet [2.7K]2 months ago
5 0
Each oxidizing agent exhibits unique characteristics tailored to various reactants. Strength varies across different solutions of oxidizing agents. For instance, sulfuric acid (H2SO4) is not as potent as nitric acid (HNO3); therefore, utilizing H2SO4 to oxidize copper is ineffective since copper requires a more robust oxidizing agent.
Alekssandra [3K]2 months ago
5 0

This appears to be an essay question based on an experiment. Different oxidizing agents possess varying strengths. When compared to nitric acid (HNO3), sulfuric acid (H2SO4) serves as a weaker oxidizing agent. Therefore, using H2SO4 in place of HNO3 would not yield the same experimental results.



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Samples of three different compounds were analyzed and the masses of each element were determined. Compound Mass N (g) Mass O (g
Anarel [2989]

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.

6 0
4 months ago
Read 2 more answers
A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man
castortr0y [3046]

Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

I hope this is useful!

7 0
3 months ago
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