Answer:
23.34 %.
Explanation:
- The percentage of water must be calculated as a mass percent.
- We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
- <em>Atomic masses of the elements:</em>
Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.
- <em>Mass of the formula unit:</em>
Cd(NO₃)₂•4H₂O
mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.
- <em> Mass of water in the formula unit:</em>
<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.
- <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
The difference between the wattage is 60-18=42W. The saving in energy over 10 hrs is 10×42=420 watt-hours=0.42kWh
Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s
All organic compounds have at least 1 carbon and 2 hydrogen atoms.