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Nastasia [14]
2 years ago
7

It takes 2 g of chlorine to sanitize 1,000,000g of water. How much chlorine will it take to sanitize a 40,000-gallon pool? ( 1 g

allon =3.7L; the density of water is 1.0g/ml ) Show your work
Chemistry
2 answers:
dimulka [17.4K]2 years ago
7 0

Answer:

0.1g (Gallon) of chlorine

Explanation:

<u>Formula</u>

1 gallon = 3.7L; the density of water is 1.0g/ml

<u>Given</u>

2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water

<u>Solve</u>

If 2g (gallon) chlorine = 1,000,000g (gallon)

∴, ? chlorine = 40,000

The First step; set up an equation

1000000/2 = 40000/?

The Next step; divide 1 million to 2

1000000 ÷ 2 = 500000

Then, divide the result by 40000

40000 ÷ 500000 = 0.08

In the nearest unit that is 0.1

Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.

riadik2000 [5.3K]2 years ago
6 0

Answer:

296

Explanation:

1 gallon = 3.7 L

40,000 gallon

= 40,000 × 3.7

= 148,000 L

1 g = 1 / 1000 L

1,000,000 g

= 1,000,000 / 1000

= 1,000 L

Let x be the amount of chlorine that will it take to sanitize a 40,000-gallon pool.

2 : 1,000 : : x : 148,000

2 / 1,000 = x / 148,000

x × 1,000 = 2 × 148,000

x = ( 2 × 148,000 ) / 1,000

= 2 × 148

x = 296 g

Therefore,

296 g of chlorine will it take to sanitize a 40,000-gallon pool.

( I solved using maths )

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8 0
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Calculate the atomic mass of silicon. The three silicon isotopes have atomic masses and relative abundances of 27.9769 amu (92.2
Kamila [148]
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6 0
3 years ago
A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of
Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0
2 years ago
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.1
Yuri [45]

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

<u>Step 1: </u>Data given

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Molarity of magnesium sulfate = 0.019 M

<u>Step 2:</u> Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

 ⇒ R = gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

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6 0
3 years ago
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Marrrta [24]
<h3><u>Answer</u>;</h3>

= 226 Liters of oxygen

<h3><u>Explanation</u>;</h3>

We use the equation;

LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;

Moles of  LiClO4;

(500 g LiClO4) / (106.3916 g LiClO4/mol)

= 4.6996 moles

Moles of oxygen;

But, for every 1 mol LiClO4, two moles of O2 are produced;

= 9.3992 moles of Oxygen

V = nRT / P

= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)

= 226 L of oxygen

5 0
3 years ago
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