Reductio is gain of electrons, while oxdation is loss. The oxidtion states of carbon in the reaction will show either of oxidation or reduction occurrence.

$C_{2} H_{2}(g) +H_{2} (g)$ → $C_{2} H_{4}(g)$

Explanation:

Electrons displacement is explained for each reaction below:

+2 to +4 = loss of 2 more lectrons-------oxidation
+4 throughout reaction----no exchange of electrons
-4 to +4 = loss of 4 electrons-----oxidation
0 to -2 = gain of 2 electrons-----reduction

5 0

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

sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(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

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

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

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

(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

5 0

4 years ago

Determine the energy of the electron in the 1s orbital of a helium ion (He+ )

Luda [366]

He Rydberg formula can be extended for use with any hydrogen-like chemical elements.
1/ λ = R*Z^2 [ 1/n1^2 - 1/n2^2] 
where 
λ is the wavelength of the light emitted in vacuum; 
R is the Rydberg constant for this element; R 1.09737x 10^7 m-1 
Z is the atomic number, for He, Z =2; 
n1 and n2 are integers such that n1 < n2 
The energy of a He+ 1s orbital is the opposite to the energy needed to ionize the electron that is 
taking it from n = 1 (1/n1^2 =1) to n2 = ∞ (1/n2^2 = 0) 
.: 1/ λ = R*Z^2 = 1.09737x 10^7*(2)^2 
λ = 2.278*10^-8 m 
E = h*c/λ 
Planck constant h = 6.626x10^-34 J s 
c = speed of light = 2.998 x 10^8 m s-1 
E = (6.626x10^-34*2.998 x 10^8)/(2.278*10^-8) = 8.72*10^-18 J ion-1 
Can convert this value to kJ mol-1: 
(8.72*10^-18*6.022 x 10^23)/1*10^3 = 5251 kJ mol-1 
Lit value: RP’s secret book: 5240.4 kJ mol-1 (difference is due to a small change in R going from H to He+) 
So energy of the 1s e- in He+ = -5251 kJ mol-1

6 0

3 years ago