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2 years ago

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What is the mass,in grams,of 1.20x10^25 atoms of calcium?

2 answers:

kow [346]2 years ago

8 0

The mass of 1.20x10^25 atoms of Ca is 8000 grams

mariarad [96]2 years ago

7 0

Answer:

The mass of calcium in the given number of atoms is 797.2 g.

Explanation:

$N=n\times N_A$

N = Number of atoms/molecules

n = Moles of compound

$N_A$ = Avogadro number = $6.022\times 10^{23} mol^{-1}$

We have:

Number calcium atoms = $N=1.20\times 10^{25}$

n = moles of calcium = ?

$n=\frac{N}{N_A}=\frac{1.20\times 10^{25}}{6.022\times 10^{23} mol^{-1}}$

n = 19.93 mol

Atomic mass of calcium = 40 g/mol

Mass of 19.93 moles of calcium = 40 g/mol × 19.93 mol=797.2 g

The mass of calcium in the given number of atoms is 797.2 g.

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3 years ago

At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

$P=X_{A}P_{A}+X_{B}P_{B}$

Here, $X_{A}$ is mole fraction of A, $X_{B}$ is mole fraction of B, $P_{A}$ is partial pressure of A and $P_{B}$ is partial pressure of B.

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$X_{A}+X_{B}=1$

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

$X_{octane}=1-X_{hexane}=1-0.58=0.42$

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

$P=X_{hexane}P_{hexane}+X_{octane}P_{octane}$

Putting the values,

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3 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago