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natali 33 [55]
3 years ago
5

The gas in a closed container has a pressure of 3.00 x 10² kPa 30 ° C. What will the pressure be if the temperature is lowered t

o -172 ° C?
Chemistry
1 answer:
Rainbow [258]3 years ago
6 0

Answer: 100kPa

Explanation:

P1 = 3.00 x 10² kPa , P2 =?

T1 = 30°C = 30 +273 = 303k

T2 = —172°C = —172 + 273 = 101k

P1/T1 = P2/T2

3 x 10² / 303 = P2 / 101

P2 = (3 x 10² / 303) x 101

P2 = 100kPa

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What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515°c/m.
Ronch [10]
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.<span>
ΔTb -  the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
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Tb(solution) = Tb(water) + ΔTb.
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4 0
3 years ago
The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature
salantis [7]

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸  J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} =   e ^{^{-\dfrac{(\Delta E)}{KT}}}

From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

k = Boltzmann constant = 1.38 × 10⁻²³  J/K

\dfrac{N_2}{N_1} = 0.5

Thus:

0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = -226811.5942 \times \dfrac{1}{T}

T =  \dfrac{-226811.5942}{-3.00 }

T = 75603.86473 K

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