I think iron because the other options are all bad conductors.
Answer:
The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.
Explanation:
Pressure in the container is P and volume is V.
Temperature of the helium gas molecules =
Molecules helium gas = x
Moles of helium has = 
PV = nRT (Ideal gas equation)
...[1]
After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.
Molecules of helium left after removal = 
Moles of helium has left after removal = 
...[2]
The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.
<h2>Ultraviolet Light</h2>
Explanation:
- The energy of a photon that will be released if an electron falls from the n= 2 orbit (excited state) to the n0 = 1 orbit (ground state) is of ultraviolet light.
- In the ultraviolet part of the spectrum, a photon having an energy of 10.2 eV has a wavelength of 1.21 x 10-7 m.
- Hence, when an electron wants to jump or it gets excited from the first level to the second level that is from n = 1 orbit to n = 2 orbits, it must absorb a photon of ultraviolet light.
- But,When an electron falls from n = 2 orbit to n = 1 orbit or from n = 2 orbit(excited state) to n = 0 orbit(groubd state), it emits a photon of ultraviolet light.
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
