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Pie
3 years ago
11

What is the mole fraction of water in pure water

Chemistry
1 answer:
irinina [24]3 years ago
5 0
In pure water, all of the molecules in the liquid are water molecules so the mole fraction is 1 (100 % H2O, 55 mol/L). In sea water, the concentration of water molecules in the solution is less than that of pure water so the vapor pressure of sea water is also lower.
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Varvara68 [4.7K]

Answer: C

creativity

Explanation:

C. Because the results of some experiments might not be explained by any

known theories.

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3 years ago
Agents that antagonize the effects of acetylcholine are called __________.
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Agents that antagonize the effects of acetylcholine are called anticholinergic. 
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3 years ago
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Write the concentration equilibrium constant expression for this reaction. cu 2 (aq) 2ag (s) → cu (s) 2ag (aq)
Llana [10]
The redox reaction given is,

                 Cu₍s₎ + 2Ag⁺₍aq₎ ---------> Cu²⁺₍aq₎ + 2Ag₍s₎

The equilibrium constant for this reaction is as follow,

                           Kc = [Cu²⁺][Ag]² / [Cu][Ag⁺]² 
 
                           Kc = [Cu²⁺] / [Ag⁺]²              ∴ {[Cu₍s₎] = 1 & [Ag₍s₎] = 1

So, Above bolded is <span>the concentration equilibrium constant expression for the given reaction.</span>
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4 years ago
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An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu
chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

The density of a proton is 6.278\times 10^{14} g/cm^3.

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\pi r^3..[1]

Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA

Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA

Volume of nucleus = V'

V=\frac{4}{3}\pi r'^3..[2]

Dividing [2] by [1]

\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}

=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}

\frac{V'}{V}=4.6656\times 10^{-14}

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm

1 m = 100 cm

Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3

Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g

1 amu = 1.66054\times 10^{-24} g

Density of the proton : d

d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3

The density of a proton is 6.278\times 10^{14} g/cm^3.

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3 years ago
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Reflection and Refraction

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