<span>Electrons are exchanged.
</span>
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
C.5
Explanation:
A number of electrons present in valence shell of penultimate Shell represents the group of elements.
For s block elements: no.of group=number of valence shell electron.
p block elements: no. of group= 2 + 10 + number of valence electrons.
d block elements: no. of group= number(n-1)d electrons + number of electrons in nth shell.
Here, the differential electron is in p orbital hence, it belongs to p block
No. of group= 2+ 10 + 3=15 i.e 15th group or VA group.
Actually, to examine it physically, then you could use a physical property that shows it's appearance. To examine its bonds etc, you'd need a chemical property. So, False.
Answer:
[OH⁻] = 8x10⁻¹⁰
Explanation:
Kw = 1x10⁻¹⁴
Kw = [H₃O⁺] . [OH⁻]
1x10⁻¹⁴ = 1.25x10⁻⁵ . [OH⁻]
1x10⁻¹⁴ / 1.25x10⁻⁵ = [OH⁻]
[OH⁻] = 8x10⁻¹⁰
