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3 years ago

24 g of magnesium to mol

Chemistry

1 answer:

sergey [27]3 years ago

7 0

Explanation:

mole = mass/ molar mass

mass= 24g

molar mass = 24g/mol

mole = 24g/ 24g/mol

= 1 mole

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3 years ago

Two unknown compounds are tested. Compound 1 contains 15.0g of hydrogen and 120.0g of oxygen. Compound 2 contains 2.0g of hydrog

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Answer:

The both compounds are different.

Explanation:

In order to confirm weather both compounds are same we will check the mole ration. If it is same the compounds will be same.

Given data:

For compound 1.

Mass of hydrogen = 15 g

Mass of oxygen = 120 g

Moles of hydrogen and oxygen = ?

Number of moles of hydrogen = 15 g/ 1g/mol = 15 mol

Number of moles of oxygen = 120 g/ 16 g/mol = 7.5 mol

Total number of moles = 22.5 mol

% of hydrogen = 15 /22.5 × 100 = 66.7%

% of oxygen = 7.5 / 22.5× 100 = 33.3%

For compound 2:

Mass of hydrogen = 2 g

Mass of oxygen = 32 g

Moles of hydrogen and oxygen = ?

Number of moles of hydrogen = 2 g/ 1g/mol = 2 mol

Number of moles of oxygen = 32 g/ 16 g/mol = 2 mol

Total number of moles = 4 mol

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3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago