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Ostrovityanka [42]
3 years ago
14

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

Chemistry
2 answers:
schepotkina [342]3 years ago
5 0

This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength.  Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.



V125BC [204]3 years ago
5 0
<h2>Answer:</h2>

Every oxidizing agent has different properties and they are specific for different reagents. Different solution of oxidizing agent will have different strength.

Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). so when you use H2SO4 for the oxidizing copper it would not work because copper need strong oxidizing agent.


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For an aqueous solution of hf, determine the van't hoff factor assuming 0% and 100% ionization, respectively. a solution is made
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According to Osmotic pressure equation:

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T= 22+273 = 295 Kelvin

So Van't half vector i = π / (MRT)
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When there is no dissociation, i = no. of moles of Hf in 1 L of solution = (1-X) 
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HF(aq) + H2O (L) ⇆ H3O (aq) + F (aq)
(1-X)                                X              X

∴ i = (1-X) + (2x)

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7 0
3 years ago
Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

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59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

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= 3.73125 moles x  -890 kJ/mol

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Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

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Hi!

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