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zubka84 [21]
3 years ago
8

A hydrogen electron is elevated from level 1 to level 2. Another electron is elevated from level 2 to level

Chemistry
1 answer:
BaLLatris [955]3 years ago
5 0

Answer: level 2 to level 4

Explanation:

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All the oxides of nitrogen have positive values of G at 298K but only one common oxide of nitrogen has a positive S identify tha
Maurinko [17]

Answer:

The common oxide of nitrogen that has a positive ΔS°f is nitric oxide (NO)

Explanation:

Without reference to thermodynamic data, we have;

1) N₂ (g) + O₂ (g) ⇄ 2 NO (g)

1 unit of N₂ + 1 unit of O₂ (total of 2 units) gives 2 units of NO, (Increase of +0 disorder)

∴ΔS°f = +ve

2) 2NO + O₂ → 2NO₂

2 unit of NO + 1 unit of O₂ (total of 3 units) gives 2 units of NO₂, (Decrease of disorder)

∴ΔS°f = -ve

3) N₂ + 1/2 O₂ → N₂O

1 unit of N₂ + 1/2 unit of O₂ (total of 1+1/2 units) gives 2 units of NO₂, (Decrease of disorder)

∴ΔS°f = -ve

4) 4 NO₂ + O₂ → 2N₂O₅

4 unit of NO₂ + 1 unit of O₂ (total of 5 units) gives 2 units of N₂O₅, (Decrease of disorder)

∴ΔS°f = -ve

5) NO + NO₂ ⇄ N₂O₃

1 unit of NO + 1 unit of NO₂ (total of 2 units) gives 1 unit of N₂O₃, (Decrease of disorder)

∴ΔS°f = -ve

Therefore, the common oxide of nitrogen that has a positive ΔS°f without reference to thermodynamic data is nitric oxide NO.

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Read 2 more answers
A compound is 2. 00% H by mass, 32. 7% S by mass, and 65. 3% O by mass. What is its empirical formula? The final step is to use
LekaFEV [45]

The empirical formula of this compound is H_2SO_4

<u>Given the following data:</u>

  • Percentage of H = 2.00%
  • Percentage of S = 32.7%
  • Percentage of O = 65.3%

<u>Scientific data:</u>

  • Molar mass of hydrogen (H) = 1.0 g/mol.
  • Molar mass of sulfur (S) = 32 g/mol.
  • Molar mass of oxygen (O) = 16 g/mol.

To determine the empirical formula of this compound:

Note: We would assume that the mass of the compound is 100 grams.

Hence, the mass of its constituent elements are:

  • Mass of hydrogen (H) = 2.00 grams
  • Mass of sulfur (S) = 32.7 grams
  • Mass of oxygen (O) = 65.3 grams

Next, we would determine the number of moles of each element by using this formula:

Number\;of\;moles = \frac{mass}{molar\;mass}

<u>For </u><u>hydrogen</u><u> (</u><u>H</u><u>):</u>

Number\;of\;moles = \frac{2.00}{1}

Number of moles = 2.0 moles

<u>For </u><u>sulfur</u><u> (</u><u>S</u><u>):</u>

Number\;of\;moles = \frac{32.7}{32}

Number of moles = 1.0 moles

<u>For </u><u>oxygen</u><u> (</u><u>O</u><u>):</u>

Number\;of\;moles = \frac{65.3}{16}

Number of moles = 4.0 moles

Empirical formula = H_2SO_4

Read more: brainly.com/question/21280037

3 0
2 years ago
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