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BartSMP [9]
3 years ago
5

PLEASE HELP FAST!! I need someone to see if my answers are correct!

Chemistry
1 answer:
11111nata11111 [884]3 years ago
5 0

Explanation:

1. The given statement is true.

Colligative properties are properties of the ideal solution which depends upon the amount or particles of solute dissolved in fixed amount of solvent. These properties do not depend upon the nature of the solute.

2. Ammonia when dissolves in water gives two particles:

NH_3+H_2O\rightarrow NH_{4}^++OH^-

3. Hexane is an non polar molecule. when dissolved in water its structural unit gives only 1 particle.

4. At higher altitudes atmospheric pressure decreases.Vapor pressure of a liquid and atmospheric pressure directly linked to each other which means that liquid will boil at lower temperature than it normal boiling temperature when boiled in  higher altitudes.

So, water will boil at temperature lower than 100° C.

5. Al(NO_3)_3 is ionic compounds which dissociated into 4 particles when dissolved in water.

Al(NO_3)_3+H_2O\rightarrow Al^{3+}+3NO_{3}^-

6.BaBr_2 is ionic compounds which dissociated into 3 particles when dissolved in water.

BaBr_2+H_2O\rightarrow Ba^{2+}+2Br^-

7. The correct answer is option A.

\Delta T_b=i\times K_b\times molality

i = van't Hoff Factor

\Delta T_b\propto i

Larger value i more will be elevation in boiling point of the solvent.

1 molecule of Ammonium phosphate dissociate into 4 particles when dissolved in water.

(NH_4)_3PO_4+H_2O\rightarrow 3NH_{4}^++PO_{4}^{3-}

8.The correct answer is option B.

\Delta T_f=i\times K_f\times molality

i = van't Hoff Factor

\Delta T_f\propto i

Larger value i more will be depression in freezing point of the solvent.

1 molecule of potassium sulfate dissociates into 3 particles when dissolved in water.

K_2SO_4+H_2O\rightarrow 2K^{+}+SO_{4}^{2-}

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Calculate the pH during the titration of 20.00 mL of 0.1000 M HNO2(aq) with 0.1000 M KOH(aq) after 13.27 mL of the base have bee
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Answer:

pH = 2.462.

Explanation:

Hello there!

In this case, according to the reaction between nitrous acid and potassium hydroxide:

HNO_2+KOH\rightarrow KNO_2+H_2O

It is possible to compute the moles of each reactant given their concentrations and volumes:

n_{HNO_2}=0.02000L*0.1000mol/L=2.000x10^{-3}mol\\\\n_{KOH}=0.1000mol/L*0.01327L=1.327x10^{-3}mol

Thus, the resulting moles of nitrous acid after the reaction are:

n_{HNO_2}=2.000x10^{-3}mol-1.327x10^{-3}mol=6.73x10^{-4}mol

So the resulting concentration considering the final volume (20.00mL+13.27mL) is:

[HNO_2]=\frac{6.73x10^{-4}mol}{0.01327L+0.02000L} =0.02023M

In such a way, we can write the ionization of this weak acid to obtain:

HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

So we can set up its equilibrium expression to obtain x as the concentration of H3O+:

Ka=\frac{[NO_2^-][H_3O^+]}{[HNO_2]}\\\\7.1x10^{-4}=\frac{x^2}{0.02023M-x}

Next, by solving for the two roots of x, we get:

x_1=-0.004161M\\\\x_2=0.003451M

Whereas the correct value is 0.003451 M. Finally, we compute the resulting pH:

pH=-log(0.003451)\\\\pH=2.462

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