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brilliants [131]

3 years ago

8

Because electron delocalization stabilizes a compound, if the only way electrons can be moved is away from the most electroneagt

ive atom, that is better than no electron delocalization at all.
a. True
b. False

1 answer:

oksano4ka [1.4K]3 years ago

5 0

It’s is false because it is

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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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