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user100 [1]
3 years ago
14

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25°C. What is the mole fraction of NaCl in this solu

tion? The vapor pressure of pure water is 23.8 torr at 25°C.
Chemistry
1 answer:
Debora [2.8K]3 years ago
3 0

Answer:

Mole fraction of Nacl is 0.173

Explanation:

we know that

P_{sol}=\chi_{solvent}P^0_{solvent}

where,

P sol  - the vapor pressure of the solution

χ  solvent - the mole fraction of the solvent

P ∘ solvent  - the vapor pressure of the pure solvent

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at  

25 ° C  You can use an online calculator to find that the vapor pressure of pure water at  25 C  is equal to about  23.8 torr .

\chi_{water}= \frac{P{sol}}{P^0{water}}

\chi_{water}= \frac{19.6}{23.8}

=0.827

Also we know that

\chi_{water}+\chi_{Nacl}= 1

This means that the mole fraction of sodium chloride is

χ_{Nacl}= 1-Χ_{water}

= 1-0.827 =0.173

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When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.
dsp73

We have that the work function of the metal

\phi=1.227*10^{-18}J

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy K.E =3.54*10-19 J.

Generally the equation for Kinetic energy    is mathematically given as

KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}

\phi=1.227*10^{-18}J

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8 0
3 years ago
I need help solving this chemistry question ​
Arturiano [62]

Answer:

I think the answer is 22.2

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3 0
3 years ago
Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia
Wittaler [7]

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

           1.375       :           2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

4 0
3 years ago
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