Answer:
X= Be
Y= B
Z=O
Explanation:
From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.
From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.
From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.
I believe it would be 1.660539040 × 10−24 gram.
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:
When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:
1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
Answer:
38 kg/m³
0.038 g/mL
Explanation:
Volume of a cube is the side length cubed.
V = s³
Given s = 0.65 m:
V = (0.65 m)³
V ≈ 0.275 m³
The mass is 10.5 kg. The density is the mass divided by volume:
ρ = (10.5 kg) / (0.275 m³)
ρ ≈ 38 kg/m³
Or:
ρ ≈ 0.038 g/mL
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
