ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants)
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) )
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 )
ΔH(reaction) = -283 - ΔH(PbO(s))
-131.4 = -283 -ΔH(PbO(s))
ΔH(PbO(s)) = -151.6 kJ
So, the best answer is A.
Answer:
The correct option is option (D).
Therefore the concentration of
is 1.8×10⁻¹⁰ M and it is basic in nature.
Explanation:
List of pH:
- If pH of a solution is 7, then solution is neutral.
- If pH of a solution is grater than 7, then solution is basic.
- If pH of a solution is less than 7, then solution is acidic.
pH of a solution is = - log₁₀[H₃O⁺]
We know that,
[H₃O⁺][OH⁻] =
=
(at 25°C)
Taking log both sides
log ([H₃O⁺][OH⁻]) =log
=log 
⇒log[H₃O⁺]+log[OH⁻] = log
= -14 log 10
[since log(mn)= log m + log n,
]
⇒ - log[H₃O⁺] - log[OH⁻] = - (-14) [ log 10 =1]
⇒pH+pOH =14.
Given that,
The concentration of OH⁻ is 5.5 × 10⁻⁵ M
[H₃O⁺][OH⁻] =
= 
⇒[H₃O⁺] 5.5 × 10⁻⁵ M=10⁻¹⁴
![\Rightarrow [H_3O^+]=\frac{10^{-14}}{5.5\times 10^{-5}}](https://tex.z-dn.net/?f=%5CRightarrow%20%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.5%5Ctimes%2010%5E%7B-5%7D%7D)
⇒[H₃O⁺] = 1.8×10⁻¹⁰ M
Now check the pH of the solution.
[H₃O⁺] = 1.8×10⁻¹⁰
Taking log both sides
log [H₃O⁺] =log( 1.8×10⁻¹⁰)
⇒ -log [H₃O⁺] = - log( 1.8×10⁻¹⁰)
⇒ pH = - log( 1.8×10⁻¹⁰)
⇒ pH =9.7.
So the nature of solution is basic.
Therefore the concentration of
is 1.8×10⁻¹⁰ M and it is basic in nature.
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>