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3 years ago

Calculate the moles of O atoms in 0.658 g of Mg(NO3)2.

Chemistry

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer:

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

Explanation:

Mass of magnesium nitrate = m = 0.658 g

Molar mass of magnesium nitrate = M = 148 g/mol

Moles of magnesium nitrate = n

$n=\frac{m}{M}=\frac{0.658 g}{148 g/mol}=0.004446 mol$

1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :

$6\times 0.004446 mol=0.026676 mol\approx 0.0267$

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

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A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

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We can calculate the moles of nitrogen using the ideal gas equation.

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3 years ago

Read 2 more answers

Using the following reaction:

Romashka-Z-Leto [24]

Answer:

Mn is the oxidizing agent.

N is the reducing agent.

Explanation:

Hello!

In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:

$Mn^{7+}+5e^-\rightarrow Mn^{2+}\\\\N^{3+}\rightarrow N^{5+}+2e^-$

Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.

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3 years ago