Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
The atomic number of the undiscovered element is 168
Element 118 will have just filled its 7p orbitals. therefore the predicted element to fill completely up to its 8 p orbital would have to filled a whole set of s, p, d, f and g orbitals
That's another 2 + 6 + 10 +14 + 18 = 50 electrons
To determine the total number of quantum numbers we have to find
Nml × Nms
we have Nml × Nms = ( 2 + 1 ) × 2
8s + 8P + 7d + 6f + 5g = 2 + 6 + 10 + 14 + 18 = 50
The element right below should be
Z = 118 + 50
= 168
Hence the atomic number of the undiscovered element is 168
Learn more about the atomic number on
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Answer:
6 electrons
Explanation:
The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s.
Answer:
The answer is partial charge 8+.
Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
