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2 years ago

A 2.0-liter aqueous solution contains a total of 3.0 moles of dissolved NH4Cl at 25°C and standard pressure.

2 answers:

vladimir1956 [14]2 years ago

7 0

The molarity is a concentration unit which defined as the number of moles of solute divided by the number of liters of solution. So the molarity of the solution is 3/2=1.5 mol/L.

bija089 [108]2 years ago

3 0

Given that

Volume of solution = 2.0 L

Number of mole of solute = 3.0 mole

Therefore molarity = number of moles / volume in L

= 3.0 Mole /2.0 L

= 1.5 mole / L

= 1.5 M

Molarity:

The number of moles of solute per liter of solution is known as molarity of the solution. It symbolized by M. Molarity of any solution is a ratio of moles of solutes per liter of solution. It expressed by the following formula:

molarity = number of moles / volume in L

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2 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

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2 years ago

Express 749 000 000 in scientific notation,

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Answer:

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What is the molar mass of: Ga2(SO3)3?

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2 years ago

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