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>
Answer:
1.17 mol
Explanation:
Step 1: Write the balanced equation
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
Step 2: Calculate the moles corresponding to 85.0 g of HCl
The molar mass of HCl is 36.46 g/mol.
85.0 g × 1 mol/36.46 g = 2.33 mol
Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl
The molar ratio of HCl to H₂ is 6:3.
2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂
Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water
Explanation:
Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.
Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm
Using above equation as:
<u>The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.</u>
Osmosis through a semipermeable membrane
I hope this helps you
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