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iragen [17]
3 years ago
15

How much heat is released when 1.4 mol of hydrogen fluoride are produced? H2(g) +F2(g) → 2HF(g) +536kJ (Please show work)

Chemistry
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

375.2 kJ

Explanation:

  • H₂(g) +F₂(g) → 2HF(g) +536kJ

The information the equation above provides lets us know that when 2 mol of hydrogen fluoride (HF) are produced, 536 kJ of energy (as heat) is produced.

We can then <u>state a rule of three</u>:

  • 2 mol HF --------  536 kJ
  • 1.4 mol HF --------- X

And <u>solve for X</u>:

  • X = 1.4 mol * 536 kJ / 2 mol
  • X = 375.2 kJ
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A balloon filled with 1.22 L of gas at 286 K is heated until the
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Explanation:

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New temperature T2 = ?

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2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

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Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
2 years ago
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