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Eduardwww [97]
3 years ago
15

How many grams of Sg is required to produce 83.10 g SF6? S: +24F-->8SF

Chemistry
1 answer:
ozzi3 years ago
7 0

Answer : The mass of S_8 required is 18.238 grams.

Explanation : Given,

Mass of SF_6 = 83.10 g

Molar mass of SF_6 = 146 g/mole

Molar mass of S_8 = 256.52 g/mole

The balanced chemical reaction is,

S_8+24F_2\rightarrow 8SF_6

First we have to determine the moles of SF_6.

\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles

Now we have to determine the moles of S_8.

From the balanced chemical reaction we conclude that,

As, 8 moles of SF_6 produced from 1 mole of S_8

So, 0.569 moles of SF_6 produced from \frac{0.569}{8}=0.0711 mole of S_8

Now we have to determine the mass of S_8.

\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8

\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g

Therefore, the mass of S_8 required is 18.238 grams.

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Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
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This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

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Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

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Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

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Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

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