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3 years ago

If Magnesium Sulfide reacts with oxygen in the air, what will be produced?

Chemistry

1 answer:

sweet [91]3 years ago

6 0

IF magnesium sulfide reacts with oxygen in the air it will produce

magnesium oxide + sulfur (IV) oxide

<u><em>explanation</em></u>

magnesium sulfide burn in oxygen to produce magnesium oxide and sulfur (iv) oxide according to the equation below

2MgS +3O2 →2MgO +2SO2

that is 2 moles of MgS react with 3 moles of O2 to produce 2 moles of MgO and 2 moles of SO2

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2 years ago

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The temp is 0.002448 of the equation

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3 years ago

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2 years ago

What mass of radon is in 8.17 moles of radon?

Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon = 8.17moles

Unknown:

Mass of radon = ?

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Number of moles = $\frac{mass}{molar mass}$

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4 0

3 years ago

Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

3 0

3 years ago