PbI(ii) ionization in the solution of PBI(ii) into water is:
<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻
If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.
Therefore,
<span>Ksp=<span>[Pb</span></span>²⁺][I-]²
Plugging the values:
1.4×10⁻⁸ = x ⋅ (2x)²
1.4×10⁻⁸ = 4x³
x³ = {1.4×10⁻⁸}÷4
x³ = 0.35 x 10⁻⁸
or
x³ = 3.5 x 10⁻⁹
x = 1.51 x 10⁻³
Hence,
Concentration of iodide ions in the solution:
2x = 3.02 x 10⁻³
Answer:
5.25g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is shown below:
Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O
From the balanced equation above,
8 moles of HF reacted to produce 2 moles of NaF.
Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.
Next, we shall convert 0.125 mole of NaF to grams.
This is illustrated below:
Mole of NaF = 0.125 mole
Molar mass of NaF = 23 + 19 = 42g/mol
Mass of NaF =..?
Mass = mole x molar mass
Mass of NaF = 0.125 x 42
Mass of NaF = 5.25g
Therefore, 5.25g of NaF is produced from the reaction.
<h3>
Answer:</h3>
28.52 seconds
<h3>
Explanation:</h3>
Initial number of atoms of Nitrogen 12,000 atoms
Half-life = 7.13
Number of atoms after decay = 750 atoms
We are required to determine the time taken for the decay.
Note that half life is the time taken for a radioactive isotope to decay to a half of its original amount.
Using the formula;
Remaining amount = Initial amount × (1/2)^n , where n is the number of half lives
In our case;
750 atoms = 12,000 atoms × (1/2)^n
0.0625 = 0.5^n
n = log 0.0625 ÷ log 0.5
n = 4
But, 1 half life =7.13 seconds
Therefore;
Time taken = 7.13 seconds × 4
= 28.52 seconds
Therefore, the time taken for 12,000 atoms of nitrogen to decay to 750 atoms is 28.52 seconds
It is purely called the Atomic Mass
Answer:
The process in which energy is emitted by one object ,transmitted through .......
