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Softa [21]
3 years ago
7

A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one par

t per million (1 ppm). (Think of 1 ppm as being 1 g chlorine per million grams of water). If you assume densities of 1.10 g/ml for the chlorine solution and 1.00 g/ml for the swimming pool water, what volume of the chlorine solution, in liters, is required to produce a chlorine level of 1 ppm in an 18,000 gal swimming pool?
Chemistry
1 answer:
deff fn [24]3 years ago
3 0

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 = \frac{\textup{68,137,470}}{\textup{ 1,000,000}}

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 = \frac{\textup{Mass}}{\textup{Density}}

or

Volume of chlorine = \frac{\textup{68.13747 g}}{\textup{1.10 g/mL}}

or

Volume of chlorine = 61.943 mL

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5 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
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Answer:

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Explanation:

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