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3 years ago

What is the empirical formula for the compound: C8H8S2?

Chemistry

1 answer:

NNADVOKAT [17]3 years ago

5 0

<h2>Hello!</h2>

The answer is:

D. $C_{4}H_{4}S$

The empirical formula of a compound is the simplest formula of the same compound, if we are given the molecular formula, we can calculate the empirical formula by finding the mole ratio of each element of the compound.

So, if we know the empirical formula we will be able to get the molecular formula by multiplying it by the mole ratio.

Then, which of the options can give as a result the given molecular formula?

The answer will be D. $C_{4}H_{4}S$ because if we multiply it by 2(mole ratio), it will give us as a result : $C_{8}H_{8}S_{2}$

$2*(C_{4}H_{4}S)=C_{8}H_{8}S_{2}$

Have a nice day!

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If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration, M, of the NaOH solution?

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3 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

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Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

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