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UkoKoshka [18]
3 years ago
9

Please help I suck at chemistry

Chemistry
2 answers:
gayaneshka [121]3 years ago
8 0
C6h5oh would be the acid
shtirl [24]3 years ago
4 0
C6H5OH donated a hydrogen to the water, so the c6h5oh would be the acid
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vladimir2022 [97]

Explanation:

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3 years ago
Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
3 years ago
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
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147.2g

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Hope this helps!

7 0
3 years ago
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