Answer:
3 years
Explanation:
Given data:
Initial amount of sample = 160 Kg
Amount left after 12 years = 10 Kg
Half life = ?
Solution:
at time zero = 160 Kg
1st half life = 160/2 = 80 kg
2nd half life = 80/2 = 40 kg
3rd half life = 40 / 2 = 20 kg
4th half life = 20 / 2 = 10 kg
Half life:
HL = elapsed time / half life
12 years / 4 = 3 years
Resources found in lithosphere: gold and iron etc
Resources found in atmosphere: Water vapor, gases etc.
Answer:
Explanation:
100mL = 0.1L
0.55 M = mol/0.1 L
mol = 0.055 mol
molar mass of KI = 165.998 g
0.055 * 165.998 = 9.13 g of KI
Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams
<u>Explanation:</u>
To calculate the mass for given number of moles, we use the equation:
Actual moles of lithium chlorate = 9.45 moles
Molar mass of lithium chlorate = 90.4 g/mol
Putting values in above equation, we get:
To calculate the theoretical yield of lithium chlorate, we use the equation:
Actual yield of lithium chlorate = 854.28 g
Percentage yield of lithium chlorate = 81.0 %
Putting values in above equation, we get:
Hence, the theoretical yield of the lithium chlorate is 1054.67 grams
