Answer:
Volume of chlorine = 61.943 mL
Explanation:
Given:
Volume of the water in the Pool = 18,000 gal
also,
1 gal = 3785.412 mL
thus,
Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL
Density of water = 1.00 g/mL
Therefore,
The mass of water in the pool = Volume × Density
or
The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g
in terms of million = 
or
= 68.13747 g
also,
1 g of chlorine is present per million grams of water
thus,
chlorine present is 68.13747 g
Now,
volume = 
or
Volume of chlorine = 
or
Volume of chlorine = 61.943 mL
Answer:
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Explanation:
Oxygen
As a trend or pattern, the atomic radius decreases as you go left to right across the period, but increases as you go down a particular group.
Answer:
B
Explanation:
u do the math and you will get the answer
Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
Learn more about standard entropy here:
brainly.com/question/14356933
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