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Hoochie [10]
3 years ago
14

What is the freezing point of an aqueous solution that boils at 105.0 ∘C? Express your answer using two significant figures.

Chemistry
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb =  ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol  . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf =  cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m  . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
4 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
Consider the following reaction at equilibrium: 2NH3 (g) N2 (g) + 3H2 (g) Le Châtelier's principle predicts that the moles of H2
Lilit [14]

Answer:

A decrease in the total volume of the reaction vessel (T constant)

Explanation:

  • Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
  • <em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
  • In this case, a decrease in volume means there is an increase in pressure, therefore the equilibrium will shift towards the side with the fewer number of moles of gas.
3 0
3 years ago
Ca+HCI—&gt;CaCI2+H2, what is the reaction?
arsen [322]

Answer:

single replacement reaction

Explanation:

This is a kind of single replacement reaction where you switch either cations or anions. Here you switched Ca for H and produced Cacl2 and H2 gas by itself.

4 0
3 years ago
Read 2 more answers
I 13 Suggest why this group is called Group 8.​
yarga [219]

Answer:

because it's in group 8

Explanation:

3 0
3 years ago
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