Which sample contains a total of 3.0 x 10^23 molecules?

1 answer:

3 0

Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:

A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.

Therefore the only choice that fits is A. 80 g of Br2.

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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

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There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$