Answer:
1.053×10²⁴ atoms of gold
Explanation:
Hello,
Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.
In this question, we're required to find the number of atoms in 344.75g of a gold nugget.
We can use mole concept relationship between Avogadro's number and molar mass.
1 mole = molar mass
Molar mass of gold = 197 g/mol
1 mole = Avogadro's number = 6.022 × 10²³ atoms
Number of mole = mass / molar mass
Mass = number of mole × molar mass
Mass = 1 × 197
Mass = 197g
197g is present in 6.022×10²³ atoms
344.75g will contain x atoms
x = (344.75 × 6.022×10²³) / 197
X = 1.053×10²⁴ atoms
Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold
3 mol KClO3 x 3 mol O2/2 mol KClO3 = 4.5 mol O2
The formula here is:
pH = -log[H+]
3.6 = -log[H+]
[H+] = 2.51 x 10⁻⁴ mol / Liter
The first option is correct.
Answer:
1.62 * 10^-11 J
Explanation:
Nitrogen-14 nucleus has seven neutrons and seven protons. So;
Total mass of protons = 7(1.00728 amu)
Total mass of neutrons = 7(1.00866 amu)
Total mass of nucleons = 7(1.00728 amu) + 7(1.00866 amu)= 14.11158 amu
Mass of Nitrogen-14 nucleus = 14.00307 amu
Mass defect = 14.11158 amu - 14.00307 amu = 0.10851 amu = 1.801852171 * 10^-28 Kg
Binding energy = Δmc^2
Where;
Δm = mass defect
c = speed of light
Binding energy = 1.801852171 * 10^-28 Kg * (3 * 10^8)^2
Binding energy = 1.62 * 10^-11 J