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vodomira [7]
3 years ago
5

A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,

if the new pressure is 1.97 atm and the new volume is 2.78 L?
Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

The answer to your question is:   T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L                    V2 = 2.78 L

P1 = 2.40 atm               P2 = 1.97 atm

T1 = 325°K                    T2 = ?

Formula

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Process

            T2 = (P2V2T1) / (P1V1)

            T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

            T2 = 1779.895 / 7.56

            T2 = 235.44 °K

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Answer:

λ = 0.5×10⁻⁵ m

E = 39.78×10⁻²¹ J

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Explanation:

Given data:

Frequency of radiation = 6.0 ×10¹³ Hz

Wavelength of radiation = ?

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Solution:

Formula:

speed of light = wavelength × frequency

3 ×10⁸ m/s = λ × 6.0 ×10¹³ Hz

Hz = s⁻¹

λ = 3 ×10⁸ m/s /  6.0 ×10¹³  s⁻¹

λ = 0.5×10⁻⁵ m

Energy of radiation:

E = hf

h = planck's constant

f = frequency

E = 6.63×10⁻³⁴ j.s × 6.0 ×10¹³  s⁻¹

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

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Answer:

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Explanation:

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