Yes, the formation of 1 mol of NH3 from H2 and N2 differ from Q for the formation of NH3 from H2 and 1mol of N2
Solution:
Equation for the formation of 1 mol of NH3
3/2H2(g)+ 1/2N2(g) → NH3(g)
The Q of this reaction is
Q1 = [product] / [reactant]
Q1 = [NH 3 ] / [H2]^3/2 [N2]^1/2
Equation for the formation of NO3 from 1 mol of N2
3H 2 (g) + N2(g) → 2NH 3(g)
The Q of this reaction
Q 2 = [product] / [reactant]
Q 2 = [NH 3 ]^2 / [H 2 ]^3 [N 2 ]
The relationship between the two Q's is
(Q1)^2 = Q2
To learn more about the Q click the given link
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<span>Electrolysis is your answer
</span>
Answer: a) %(C/Co) = (e^(-0.027t)) × 100
b) t1/2 = 25.67years
c) 5.872%
Explanation:
a) Radioactive reactions always follow a first order reaction dynamic
Let the initial concentration of Strontium-90 be Co and the concentration at any time be C
The rate of decay will be given as:
(dC/dt) = -KC (Minus sign because it's a rate of reduction)
The question provides K = 2.7% per year = 0.027/year
(dC/dt) = -0.027C
(dC/C) = -0.027dt
∫ (dC/C) = -0.027 ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from Co to C and the Right hand side from 0 to t.
We obtain
In (C/Co) = -0.027t
(C/Co) = (e^(-0.027t))
In percentage, %(C/Co) = (e^(-0.027t)) × 100
(Solved)
b) Half life of a first order reaction (t1/2) = (In 2)/K
K = 0.027/year
t1/2 = (In 2)/0.027 = 25.67 years
c) percentage that remains after 105years,
%(C/Co) = (e^(-0.027t)) × 100
t = 105
%(C/Co) = (e^(-0.027 × 105)) × 100 = 5.87%
Answer: -
First Ionization energy IE 1 for element X = 801
Here X is told to be in the third period.
So principal quantum number n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom. Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
Answer : The volume of 
is 14.784 L.
Solution : Given,
Mass of Aluminium = 6 g
Molar mass of Aluminium = 27 g/mole
First we have to calculate the moles of aluminium.
Moles of Al = 
The given balanced reaction is,
From the reaction, we conclude that
2 moles of Al react with the 6 moles of
0.22 moles of Al react with 
of
At STP, 1 mole contains 22.4 L volume
As, 1 mole of contains 22.4 L volume of
0.66 moles of contains 
volume of
Therefore, the volume of is 14.784 L.
