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3 years ago

Which of the following can be used to neutralize an ammonia (NH3) solution?

2 answers:

8 0

It would have been much better if you had attached some options to choose. But I think I've got what you mean. I bet that solution is HCl.

tensa zangetsu [6.8K]3 years ago

5 0

It really depends on your options.
Since NH3 solution is a weak base, the option that neutralizes a base is an acid. So find an acid among your options, and here are some examples: HNO3, HCl, etc.

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The Answer Is

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2 years ago

What is the pressure of 4 moles of helium in a 50 L tank at 308k using PV=nRT

Ilya [14]

Hey there!:

p = ??

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3 years ago

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2 years ago

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Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

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

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

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

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

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2 years ago