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Murljashka [212]
2 years ago
6

what is the change in freezing point of a 1.75 molal solution of nacl in water? the kf for water is 1.86 degrees c/m.

Chemistry
2 answers:
Savatey [412]2 years ago
5 0

Answer:

As the molality increases by 1 m, there is a corresponding decrease of 1.86°C in the solution’s freezing point.

Explanation:

Anser on EDge

ella [17]2 years ago
4 0

Answer:

-6.51

Explanation:

Multiply 1.86 x 2 x 1.74 and then you get your answer.

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How many kj of heat are needed to completely melt 32.3 g of h2o, given that the water is at its melting point? the heat of fusio
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Answer: fourth option, 10.8 kJ


Explanation:


The <em>heat of fusion</em>, also named latent heat of fusion, is the amount of heat energy required to change the state of a substance from solid to liquid (at constant pressure).


The data of the <em>heat of fusions</em> of the substances are reported in tables and they can be shown either per mole or per gram of substance.


In this case we have that the<em> heat of fusion for water </em>is reported per mole: <em>6.02 kJ/mole</em>.


The formula to calculate <em>how many kJ of heat (total heat) are needed to completely melt 32.3 g of water, given that the water is at its melting point</em> is:

  • Heat = number of moles × heat of fusion

The calculations are:

  • number of moles = mass / molar mass

        number of moles = 32.3 g / 18.015 g/mol = 1.79 mol

       

  • Heat = 1.79 mol × 6.02 kJ / mol = 10.8 kJ ← answer
5 0
3 years ago
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The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

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At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

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ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

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