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shepuryov [24]
3 years ago
7

Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl

orine per million grams of water. calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 9.00 percent chlorine by mass and there are 2.29 × 104 gallons (gal) of water in the pool (1 gal = 3.79 l; density of liquids = 1.00 g/ml). enter your answer in scientific notation.
Chemistry
1 answer:
frez [133]3 years ago
6 0

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is 2.29\times 10^{4} gal.

Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

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Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

3 0
3 years ago
What is the name of Bel on the periodic table
rewona [7]

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2 years ago
A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and
wariber [46]

Answer:

5

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.

3 0
3 years ago
When KNO3 dissolved in water, what is the intermolecular attraction between NO3 ions
Snezhnost [94]
Ion-dipole forces

H2O has hydrogen bonding, which is a form of dipole-dipole forces, and NO3- is an ion, so the intermolecular attraction is ion-dipole.
4 0
2 years ago
PLEASE HELP
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Answer:

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