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3 years ago

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How many grams of N2 are required to react with 2.30 moles of Mg in the following process? 3 Mg + N2 → Mg3N2? (Mg = 24.3 g/mol,

N = 14.0 g/mol)

1 answer:

aivan3 [116]3 years ago

5 0

Answer: 10.73 g

Explanation:

2.3 mol Mg * (1 mol N2 / 3 mol Mg) * (14 g Mg / 1 mol g) = 10.73

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Suppose that you melt 24 ml of ice. What is the volume of liquid water that results?

serious [3.7K]

The answer to this would be 22 mL

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3 years ago

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

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3 years ago

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Answer:

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Explanation:

Temperature decreases with an increase in altitude. At sufficiently high altitudes, the air is much colder than the air on the surface of the Earth. Water vapor has a low density and is capable to rise until it gets into a medium with a significantly low temperature and pressure to condense and produce water droplets out of water vapor. Those droplets then form clouds.

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3 years ago

What frequency corresponds to an absorption line at 460 nm?

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Answer:

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3 years ago

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4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.

Elodia [21]

Answer:

$Y=48.6\%$

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

$2Cu+O_2\rightarrow 2CuO$

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

$m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO} =0.078gCuO$

Therefore, the percent yield is:

$Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%$

Best regards.

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3 years ago