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Nonamiya [84]
3 years ago
12

Calculate the mass percent for all components in a solution containing the following. 0.350 Kg of water, 5.4 moles of ammonia an

d 195.0 grams of cobalt (II) nitrate. (The salt completely dissociates)
Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer:

  • % Water = 54.96%
  • % Ammonia = 0.14%
  • % Cobalt (II) Nitrate = 30.62%

Explanation:

To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:

  • Mass Water ⇒ 0.350 kg Water = 350 g water
  • Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol =  91.8 g
  • Mass cobalt (II) nitrate ⇒ 195.0 g

Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate

  • Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g

To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:

  • % Water ⇒ 350/636.8 * 100% = 54.96%
  • % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
  • % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
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Read 2 more answers
1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in
FrozenT [24]

Answer:

1)  <u>10.0 moles of NO</u>

<u>2) 25 moles of NaCl</u>

3) <u>1200 moles of CO2</u>

<u>4) 1.03 moles of MgO</u>

<u>5) 0.72 moles H2</u>

<u>6) 1041.15 grams BaCl2</u>

<u>7) </u>9.55 grams MgO

8) <u>45.5 grams Au</u>

<u>9 )14.93 grams AlCl3</u>

Explanation:

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO

For 5.0 moles of N2 we need 5.0 moles of O2 to produce <u>10.0 moles of NO</u>

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with  NaOH how many moles of NaCl will be formed?

HCl + NaOH → NaCl + H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce <u>25 moles of NaCl</u> and 25 moles of H2O

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and  burn 150 moles of octane how many moles of carbon dioxide are you producing?

2C8H18 + 25O2 → 16CO2 + 18H2O

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O

For 150 moles of octane we need 25*75 = 1875 moles of O2

To produce 16*75 = <u>1200 moles of CO2</u> and 18*75= 1350 moles

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how  many moles of magnesium oxide will be formed?

2Mg + O2 → 2MgO

Moles of Mg = 25.0 g/24.3 g/mol = 1.03 moles

For 2 moles Mg we need 1 mol O2 to produce 2 moles MgO

For 1.03 moles Mg we'll have <u>1.03 moles of MgO</u>

<u />

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5. . Lithium reacts with water in a single replacement reaction. How many moles of  hydrogen gas a produced by 10 grams of lithium?

2Li + 2H2O → 2LiOH + H2

Moles Li = 10.0 grams/ 6.94 g/mol = 1.44 moles

For 2 moles Li we need 2 mole H2O to produce 2 moles LiOH and 1 mol H2

For 1.44 moles Li we need 1.44 moles H2O to produce 1.44 moles H2O and <u>0.72 moles H2</u>

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6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How  many grams of barium chloride are required to react with 5 moles of sodium sulfate?

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

For 1 mol of BaCl2 we need 1 mol of Na2SO4 to produce 1 mol of BaSO4 and 2 moles NaCl

For 5 moles Na2SO4 we need 5 moles BaCl2

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7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of  magnesium carbonate are heated?

MgCO3 → MgO + CO2

Moles MgCO3 = 20.0 grams / 84.31 g/mol

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For 1 mol MgCO3 we'll have 1 mol MgO and 1 mol CO2

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8. If 70 grams of gold III chloride decomposes into its elements, how many grams of  gold will be produced?

2AuCl3 → 2Au + 3Cl2

Moles AuCl3 = 70 grams / 303.33 g/mol = 0.231 moles

For 2 moles AuCl3 we'll have 2 moles gold and 3 moles Cl2

For 0.231 moles AuCl3 we'll have 0.231 moles gold

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9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2

Moles AlBr3 = 30 g /266.69 g/mol = 0.112 moles

For 2 moles AlBr2 we need 3 moles Cl2 to produce 2 moles AlCl3 and 3 moles Br2

For 0.112 moles AlBr3 we need 3/2 * 0.112 = 0.168 moles of Cl2

To produce 0.112 moles of AlCl3 and 0.168 moles of Br2

Mass AlCl3 = 0.112 moles * 133.34 g/mol = <u>14.93 grams AlCl3</u>

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3 years ago
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Galina-37 [17]

The standard enthalpy of formation for chlorine is zero but the standard entropy is larger than 0 because it is the elemental state of chlorine.

The standard enthalpy of formation for chlorine is zero because cl2 is the elemental state of chlorine and it does not require any energy for the formation of the standard state of chlorine.

The entropy of any system cannot be negative. It can only be positive or zero.

The entropy of a system will become zero only at a absolute zero temperature.

That's why the entropy of chlorine in elemental state is more than zero because absolutely zero temperature can't be obtained.

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7 0
1 year ago
what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4
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Answer:

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First, find the hydronium ion concentration of the solution with a pH of 4.

[H₃O⁺] = 10^-pH

[H₃O⁺] = 10⁻⁴

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Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.

[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01

Lastly, find the pH.

pH = -log [H₃O⁺]

pH = -log (0.01)

pH = 2

The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.

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4 0
3 years ago
Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17
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n(C</span>₄H₁₀) = 0,8 mol ÷ 4.<span>
n(C</span>₄H₁₀) = 0,2 mol.<span>
m(C</span>₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).<span>
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m(C</span>₄H₁₀) = 11,6 g.

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