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4 years ago

Which quantity will correctly convert between two different substances

Chemistry

1 answer:

DENIUS [597]4 years ago

4 0

Molar ratio is the quantity which will correctly convert between two different substances.

Answer: Option B

<u>Explanation: </u>

Molar ratio is used to determine the number of moles of reactants required to convert to desired number of moles of the products and vice versa. As moles are the coefficient written in any chemical equation. The ratio of the number of moles of the reactants to the number of moles of the products will help to determine the molar ratio.

This molar ratio is used as the conversion factor to determine number of moles required in the reactant side for the reactants to form a desired product of desired moles. We can also determine the number of moles utilized from the reactants from the moles obtained in the products using this molar ratio conversion factor.

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3 years ago

A compound has a molecular weight of 146 g/mol. A 0.3250 g sample of the compound contains 0.1605 g of carbon, 0.0220 g of hydro

Sonbull [250]

Answer: The molecular formula is $C_6H_{10}S_2$

Explanation:

We are given:

Mass of $C$ = 0.1605 g

Mass of $H$ = 0.0220 g

mass of $S$ = 0.1425 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.1605g}{12g/mole}=0.0134moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0220g}{1g/mole}=0.0220moles$

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.1425g}{32g/mole}=0.0044moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.0134}{0.0044}=3$

For H = $\frac{0.0220}{0.0044}=5$

For S = $\frac{0.0044}{0.0044}=1$

The ratio of C : H: S= 3: 5: 1

Hence the empirical formula is $C_3H_5S$

The empirical weight of $C_3H_5S$ = 3(12)+5(1)+1(32)= 73g.

The molecular weight = 146 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{146}{73}=2$

The molecular formula will be= $2\times C_3H_5S=C_6H_{10}S_2$

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3 years ago

An aqueous solution contains 0.35 m ammonium nitrate. one liter of this solution could be converted into a buffer by the additio

LuckyWell [14K]

Answer:

A strong base, such as NaOH. The amount of OH added shouldn't exceed 0.35 mols (though i would stop at 0.30 mols)

Explanation:

a weakly basic salt can be turned into a buffer by the addition of a strong base, and a weakly acidic salt can be turned into a buffer with a strong acid

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1 year ago