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Setler79 [48]
3 years ago
11

Will give 20 points and brainliest help

Chemistry
1 answer:
Blababa [14]3 years ago
8 0

the answer you have chosen is correct

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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
I NEED HELP ASAP PLZ LOOK AT THE PICTURE CAREFULLY!!!!
Virty [35]

Answer:

it is a little fuzzy

Explanation:

6 0
3 years ago
Helen recorded the following data about the half-life of a radioisotope. Radioactive Decay of Radioisotope A Grams of Radioactiv
MissTica

Answer:

Line graph

Explanation:

I did it on Study Island

7 0
3 years ago
Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
3 years ago
If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0
3 years ago
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