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Georgia [21]
2 years ago
5

The freezing point of pure chloroform is -63.5°C, and its freezing point depression constant is 4.07°C•kg/mol. If the freezing p

oint of a solution of benzoic acid in chloroform is -70.55°C, what is the molality of this solution?
Chemistry
1 answer:
Nady [450]2 years ago
4 0

Answer:

-20.22

Explanation:

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2 days Claire went missing, the body of a young girl was found in an abandoned building. There were 2 blowfly larvae on the body
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Yes it is possible that the body might be Claire's

Explanation:

It is possible because the scenario says that it had similar appearances that Claire had

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What is the reaction called when an acid cancels with a base?
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Mixing a base with an acid results in a chemical reaction called neutralization.<span> The result is a perfectly balanced solution of salt and water with a pH of 7 if the acid and base are balanced properly. Depending on the bases and acids used, it can be a dangerous experiment.</span>
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Density equals what?
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2 years ago
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A solution is prepared by dissolving 0.56 g of benzoic acid (C6H5CO2H, Ka ???? 6.4 ???? 10????5) in enough water to make 1.0 L o
garri49 [273]

Answer:

[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22

Explanation:

Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵

M (molar mass) of BA (Benzoic Acid) = 122 g/mol

Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M

We should consider the equation once it reaches the equilibrium:

C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺

  C - x                      x              x

And, for the Kₐ:

Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³

Then: x² + Kₐx - KₐC = 0

x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0

Resolving this cuadratic equation (remember to use Baskara equation), we obtain:

x = 6.083x10⁻⁴ M

Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M

[C₆H₅COOH] = C - x = 3.98x10⁻³ M

pH = -Log [H⁺] = 3.22

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3 years ago
Which one of the following is a free radical?
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Answer:

where are the following

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