ANSWER ASAP! PLEASE!

kirza4 [7]

2)8/8-9_12-/78)022-/7/$

6 0

2 years ago

For the solution resulting from dissolved 0.32 g of naphthalene (C10H8) in 25 g of benzene (C6H6) at temperature of 26.1°C, calc

grandymaker [24]

Answer:

See explanation

Explanation:

Number of moles of naphthalene = 0.32g/128.1705 g/mol = 0.0025 moles

Molality = number of moles/ mass of Solvent in kilograms

Molality = 0.0025/0.025 Kg

Morality = 0.1 m

But

∆T= K × i × m

Where ∆T = boiling point elevation

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= boiling point elevation constant = 5.12 °C/m

∆T= 5.12 °C/m ×0.1 = 0.512°C

For freezing point depression

∆T= K× i × m

Where ∆T= freezing point depression

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= freezing point depression constant = 2.67 °C/m

∆T= 2.67 °C/m ×0.1 = 0.267°C

From Raoult's law;

∆P = XBPA°

Where;

∆P = vapour pressure lowering

XB = mole fraction of solute

PA° = vapour pressure of pure solvent

Number of moles of solvent = mass/molar mass = 25g/ 78 g/mol= 0.3205 moles

Total number of moles = number of moles of solute + number of moles of solvent = 0.0025 moles + 0.3205 moles = 0.323 moles

Mole fraction of solute = 0.0025 moles/0.323 moles = 0.0077

Vapour pressure of benzene = 100 torr

Therefore;

∆P = 0.0077 × 100torr = 0.77 torr

Hence;

∆P = 0.77 torr

5 0

2 years ago

How much force is required to lift a rock a vertical distance of 8 m if 80 j of work were done

yanalaym [24]

we have

work done (W)= force(F) × displacemen(s)

or, 80= F× 8

or, F= 10 N

therefore, 10 N force is required to lift the rock.

5 0

2 years ago

The kb of hypochlorous acid is 3. 0×10^–8 at 26. 0 C. What is the percent of ionization of hypochlorous?

astraxan [27]

The acid dissociation constant (Ka) defines the difference between a weak and a strong acid. The % ionization of hypochlorous acid is 0.14%.

<h3>What is the acid dissociation constant?</h3>

The acid dissociation constant is used to define the ionization constant of an acidic substance. It gives the quantitative measurement of the strength.

The ICE table is attached to the image below.

The acid dissociation constant (Ka) for the reaction is,

Ka = [H⁺][ClO⁻] ÷ [HClO]

= a² ÷ (0.015 - a)

= 3.0 x 10⁻⁸

Now, a² + 3.0 x 10⁻⁸ a - 4.5 × 10⁻¹⁰ = 0

So, a = 2.210 × 10⁻⁵

Solving further,

[H+] = a = 2.210 × 10⁻⁵ M

The percent ionization is calculated as,

[H+] ÷ [HClO] × 100

= 2.210 × 10⁻⁵ M ÷ 0.015 × 100

= 0.14 %

Therefore, 0.14 % is the percentage of hypochlorous ionization.

Learn more about acid dissociation constant here:

brainly.com/question/22668939

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Your question is incomplete, but most probably your full question was, The ka of hypochlorous acid (HClO) is 3.0 x 10⁻⁸ at 25.0°C. What is the % of ionization of hypochlorous acid in a 0.015 aqueous solution of HClO at 25.0C?