When considering atomic orbitals the only important information they really wanted to know is the size of the orbit, which was described by using quantum numbers.
C.... 376 tbh just took the test and was right
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M
Answer:
the sum is Rs 4,200
Explanation:
Given that
3 by 7 of the sum is $1,800
We need to find out the sum
So,
Here we assume the sum be x
So the following equation should be made
3 ÷ 7 × x = 1800
3x = 1800 × 7
x = (1800 × 7) ÷ 3
= Rs 4,200
Hence, the sum is Rs 4,200
The same would be considered and relevant
