Answer:
Ca(OH)2 will not precipitate because Q<Ksp
Explanation:
Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6
The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.
The reaction equation is:
CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂
From
Ca(OH)2= Ca2+ + 2OH-
Concentration of solution= 0.064×1/64= 1×10-3
Since [Ca2+] = 1×10-3
[OH-]= (2×10-3)^2= 4×10^-6
Hence Q= 4×10^-9
This is less than the Ksp hence the answer.
Answer:
0.416 mol CaBr₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
83.1 g CaBr₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar mass of Br - 79.90 g/mol
Molar Mass of CaBr₂ - 40.08 + 2(79.90) = 199.88 g/mol
<u>Step 3: Convert</u>
<u />
= 0.415749 mol CaBr₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.415749 mol CaBr₂ ≈ 0.416 mol CaBr₂
49 neutrons in each nucleus.
<h3>Explanation</h3>
For each nucleus:
Mass number = Number of protons + Number of neutrons.
The atomic number of a nucleus is the same as its number of protons. The atomic number of the nucleus here is 31. There are 31 protons in each nucleus.
- Mass number = 80;
- Number of protons = Atomic number = 31;
- The number of neutrons is to be found.
Again,
Mass number = Number of protons + Number of neutrons.
80 = 31 + Number of neutrons.
Number of neutrons = 80 - 31 = 49.
PH is defined as the negative log of Hydrogen ion concentration. Mathematically we can write this as:
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
We are given the concentration of

. Using the value in formula, we get:
Therefore, the pH of the solution will be 3.745
Answer:
Where Blocal = local magnetic field between the two regions of the molecule
Blocal = (1-σ)B0
ΔBlocal = (1-σ1)B0 - (1-σ2)B0 = (σ2 - σ1)B0 = ΔσB0 ≈ ΔδB0 x 10∧-6
= (3.36-1.16) x 10∧-6 x B0 = 2.20 x 10∧-6B0
(a) ΔBlocal = 2.20 x 10∧-6 x 1.9T = 4.2 μT
(b) ΔBlocal = 2.20 x 10∧-6 x 16.5T = 36.3 μT
Explanation: