Someone help me please. It's due in 30 mins

What is the boiling point of an aqueuous solution of

Olegator [25]

Answer:

100.223°C is the boiling point of an aqueous solution.

Explanation:

Osmotic pressure of the solution = π = 10.50 atm

Temperature of the solution =T= 25 °C = 298 .15 K

Concentration of the solution = c

van'y Hoff factor = i = 1 (non electrolyte)

$\pi =icRT$

$c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}$

c = 0.429 mol/L = 0.429 mol/kg = m

(density of solution is the same as pure water)

m = molality of the solution

Elevation in boiling point = $\Delta T_b$

$\Delta T_b=iK_b\times m$

$\Delta T_b=T_b-T$

T = Boiling point of the pure solvent

$T_b$ = boiling point of the solution

$K_b$ = Molal elevation constant

We have :

$K_b=0.52^oC/m$ (given)

m = 0.429 mol/kg

T = 100° C (water)

$\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg$

$\Delta T_b=0.223^oC$

$\Delta T_b=T_b-T$

$T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC$

100.223°C is the boiling point of an aqueous solution.

3 0

3 years ago

How much hbro must be added to 1l of pure water to make a solution with a ph of 4.25? ka = 2.00 × 10−9 ?

RUDIKE [14]

Answer is: 153.52 grams of hypobromous acid <span>must be added.
</span>Chemical dissociation: HBrO ⇄ H⁺ + BrO⁻.
pH = 4.25.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-4.25).
[H⁺] = [BrO⁻] = 5.62·10⁻⁵ M.
Ka = [H⁺] · [BrO⁻] / [HBrO].
2.00·10⁻⁹ = (5.62·10⁻⁵ M)² / [HBrO].
[HBrO] = 3.16·10⁻⁹ M² / 2.00·10⁻⁹.
[HBrO] = 1.58 M.
m(HBrO) = n(HBrO) · M(HBrO).
m(HBrO) = 1.58 mol · 96.91 g/mol.
m(HBrO) = 153.52 g.

3 0

4 years ago

A ball is thrown upward as shown. Gravity pulls (accelerates) the ball back down. Which of the following statements is true?

liraira [26]

Answer:

I think third one is the right answer

Explanation:

if it is right please mark me as brainliest

4 0

3 years ago

Read 2 more answers

What does the rate law use to determine the rate of a reaction?

Alexandra [31]

Answer:

B

Explanation:

reaction order is simply the sum of orders for each reactant.

3 0

3 years ago

How would you make 5mls of a solution that is 2.0% lactose and 0.1M SPG buffer from separate stock solutions that are 6% lactose

Aliun [14]

Answer:

Check the explanation section.

Explanation:

The following steps should be followed orderly.

STEP ONE:

Use the dilution equation in the calculation of the volume for the stock solution. That is, C1 × V1 = C2 × V2.

Where C1 and C2 are the concentration of the stock solution and the diluted solution.

STEP TWO:

Put 6% of lactose and make sure to dilute it in order to make 2.0% lactose

and put it in Beaker A. Also, make sure to dilute the 1M to 0.1M SPG buffer in Beaker B.

STEP TWO:

Now, from beaker A containing 2% lactose, measure and remove 5.0 mL from it. Also, measure and remove 5.0 mL from beaker B containing 0.1M SPG.

So, in STEP TWO above we won't know how much water we need to use for dilution, thus, there is the need to make use of STEP ONE.

Therefore, from STEP ONE ABOVE, we have the dilution equation given as;

C1 × V1 = C2 × V2.

Hence, 6 × V1 = 2 × 5. Therefore, the volume needed from the stock solution, V1 = (2 × 5)/ 6 = 1.6 mL.

STEP THREE:

Now measure out 1.6 mL from the stock solution, that is 6% lactose and add it to 5mL of the diluted solution of 2% in beaker A into another container, say beaker C and add H2O to form SOLUTION X.

STEP FOUR:

Using the dilution equation again, Determine the the volume that is needed from 1M SPG.

C1 × V1 = C2 × V2.

V1 = ( 0.1 × 5)/ 1 = 0.5mL.

STEP FIVE:

measure 0.5mL out from the 1M SPG and 5 mL out of 0.1M SPG buffer and add water to it to form SOLUTION Y.

STEP SIX:

Now, mix solution X and solution Y together and take the required 5ml

7 0

3 years ago