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Yuliya22 [10]
3 years ago
5

A one electron species, X m, where m is the charge of the one electron species and X is the element symbol, loses its one electr

on from its ground state when it absorbs 3.49 x 10-17 J of energy. Using the prior information, the charge of the one electron species is:_____________
a. +8
b. +2
c. +3
d. +1
e. +4
Chemistry
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

Option C

Explanation:

From the question we are told that:

Difference in energy \delta E =3.49 * 10^{-17} J

The Ground state Difference in energy at n=1

\delta E_g = 2.18 * 10^{-18} × Z^2

Generally the equation for Difference in energy is mathematically given by

\delta E=\delta E_g

Therefore

3.49 * 10^{-17} = 2.18 * 10^{-18} * Z^2

Z^2=16

Z=4

Therefore

Charge on element Z Q_Z

Q_Z= Atomic\ no\. of\ element - No.\ of\ electrons\ of\ element

Q_Z =4-1

Q_Z=+3

Option C

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3 years ago
Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin
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Answer:

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(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

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Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

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Mass = 3.171 kg

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or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

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