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PIT_PIT [208]
3 years ago
5

On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential

energy of two O atoms versus the distance between their nuclei.

Chemistry
1 answer:
ki77a [65]3 years ago
6 0

Answer:

Explanation:

We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.

From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.

Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.

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What mass of AI2O3 forms from 16 g O2 and excess AI?
vlada-n [284]

Answer:

2Al+1.5O2→Al2O3

Thus, 2 mol of Al combine with 1.5 mol of oxygen to form 1 mol of Al2O3.

2 mol of Al corresponds to 2×27=54 g.

Thus, the weight of Al used in the reaction is 54 g.

4 0
2 years ago
If 224 g of iron reacts with oxygen, forming
Likurg_2 [28]

Answer: 208g

Explanation: Since 224 grams produces 96 grams more, then 112 grams will produce 208 grams.

8 0
3 years ago
Consider the work of Thomson, Rutherford, and Bohr. In your opinion, which was the most important and why? Be sure to state what
solong [7]
Rutherford found protons Thompson found electrons and Bohr crewed the Bohr model the most commen model of the atom used today This really a your opinion question but my favor is Bohr
3 0
3 years ago
What’s the answer??????
atroni [7]

Answer:

A evaperation

Explanation:

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7 0
3 years ago
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
3 years ago
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