Yes it need to be like that cause when it like that it like that
Answer:
What do <u>YOU</u> think?
Explanation:
This question is asking for an opinion. The word should is included.
Answer:
Explanation:
a ) false.
NH₃ is more polar molecule than PH₃ so inter-molucular attraction is greater in NH₃ ( hydrogen bond ) . Hence vapour pressure is low for NH₃ .
b ) false .
The average kinetic energy of boiling water molecules is lower on a mountaintop than it is at sea level. It is so because water boils at lower temperture on mountain and kinetic energy of molecules depends upon temperature .
c ) false
vapour pressure depends upon temperature .
d ) True
CCl4 is more volatile than CBr4
e ) false
vapour pressure increases as temperature increases.
Answer:
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. ( B )
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not. ( C )
Explanation:
<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. and
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.
Given the relationship below,
Δ G = -nFE
E = emf of cell , G = free energy.
This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
