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Svetllana [295]
3 years ago
10

Question 1

Chemistry
2 answers:
Verizon [17]3 years ago
6 0

Answers are in document attached.

Download docx
Nesterboy [21]3 years ago
3 0
P waves are faster S wave are slower P waves can pass through both solids and liquids S waves bounce of solids
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What is reproduction
Zolol [24]
The action or process of making a copy of something.
3 0
3 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
Read 2 more answers
Which power source uses the earth’s internal heat?
Nataly_w [17]

Answer:

Its geothermal energy.

Explanation:

hope this helps

5 0
3 years ago
A bird sits on her eggs to keep them warm
Troyanec [42]

Answer:

This would be Conduction

Explanation:

Conduction is the transfer of heat between two surfaces in contact with each other like the bird and the eggs.

5 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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