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Ede4ka [16]
2 years ago
12

What is the difference between a pure solvent and a solution? Do they have the same physical properties?

Chemistry
1 answer:
pychu [463]2 years ago
5 0

Answer:

The physical properties of a solution are different from those of the pure solvent. ... Colligative properties are those physical properties of solutions of nonvolatile solutes that depend only on the number of particles present in a given amount of solution, not on the nature of those particles.

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What is the pH of a solution of KOH with a hydroxide concentration of [OH⁻] = 1.10 x 10⁻⁴
olga2289 [7]

Answer:

p[H+] = 10.042

Explanation:

As we know that

pKw = pH + pOH......eq (1)

we will calculate the pH of OH- and then we will calculate the pH of H+

So p[OH-] = - log [1.10 * 10^{-4}]

Solving the right side of the equation, we get

p[OH-]

= - [-3.958]\\= 3.958

Now we know that

pKw = 14.0

Substituting the value of pOH in the above equation, we get -

14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042

3 0
3 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
3 years ago
The Great Burdock plant’s seeds have spines on them that attach to the fur of animals that brush against it. The seed then trave
Sphinxa [80]
The answer is B. Commensalism.

<span>Commensalism is a relationship between two organisms in which only one of them has benefit, and the other one is not affected. In this example, the Great Burdock's plants spread their seeds using animals, so they benefit from this relationship. On the other hand, animals neither have benefits not are harmed from the relationship.</span>
3 0
3 years ago
Read 2 more answers
Which of the following was not a big ideal of chemistry
ikadub [295]

Answer:

I think the answer is gravity

3 0
2 years ago
How many liters of radon gas would be in 3.43 moles at standard temperature and pressure (273 K and 100 kPa)?
MissTica

Answer: Option B. 76.83L

Explanation:

1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.

If 1 mole of Radon = 22.4L

Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L

5 0
3 years ago
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