Explanation:
To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.
Based on this premise, the number of moles of species on the reactant and product side must be the same;
Li + Br₂ → LiBr
Put a,b and c as the coefficient of each species
aLi + bBr₂ → cLiBr
balancing Li;
a = c
balancing Br;
2b = c
let a = 1;
c = 1
b =
or a = 2, b = 1 , c = 2
2Li + Br₂ → 2LiBr
P + Cl₂ → PCl₃
Using the same method;
aP + bCl₂ → cPCl₃
balancing P;
a = c
balancing Cl;
2b = 3c
let a = 1;
c = 1
b =
or
a = 2, b = 3, c = 2
2P + 3Cl₂ → 2PCl₃
iii,
H₂ + SO₂ → H₂S + H₂O
use coefficients a,b,c and d;
aH₂ + bSO₂ → cH₂S + dH₂O
balancing H;
2a = 2c + 2d
balancing S;
b = c
balancing O
2b = d
let b = 1,
c = 1
d = 2
a = 3
3H₂ + SO₂ → H₂S + 2H₂O
Answer:
c.boron-11
Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.81u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u=ux
x=0.19
1−x=0.81
And thus the abundance of boron-11 is roughly 81%.
The answer to your question is
D: Unshared Pair
Hope this helps you :))))
10 is mixture of elements and compounds
11 is compound
12-15 are mixture of compounds