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bearhunter [10]
3 years ago
7

Tin(II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in

33.7 g of the compound?
Chemistry
1 answer:
Alexeev081 [22]3 years ago
6 0

Answer:

In 33.7 grams SnF2 we have 8.17 grams of F

Explanation:

Step 1: Data given

Mass of SnF2 = 33.7 grams

Molar mass of SnF2 = 156.69 g/mol

Molar mass of F = 19.00 g/mol

Step 2: Calculate moles of SnF2

Moles SnF2 = mass / molar mass

Moles SnF2 = 33.7 grams / 156.69 g/mol

Moles SnF2 = 0.215 moles

Step 3: Calculate moles F

For 1 mol SnF2 we have 2 moles F

For 0.215 moles SnF2 we have 2*0.215 = 0.430 moles F

Step 4: Calculate mass F

Mass F = moles F * molar mass F

Mass F = 0.430 moles * 19.00 g/mol

Mass F = 8.17 grams

In 33.7 grams SnF2 we have 8.17 grams of F

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Answer:

0.846 moles.

Explanation:

  • This is a stichiometric problem.
  • The balanced equation of complete combustion of butane is:

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

  • It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.

<u><em>Using cross multiplication:</em></u>

  • 1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂
  • ??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂
  • The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.
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If a sample is thought to be several million years old, which method would best help to determine its absolute age?
Makovka662 [10]
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I have two solutions. In the first solution, 1.0 moles of sodium chloride is dissolved to make 1.0 liters of solution. In the se
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i not sure but I think yes

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2 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
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