Answer:
38.7%
41.3%
20%
Explanation:
The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.
To solve this problem:
- find the formula mass by adding the atomic masses of the atoms that makes up the compound.
- place the mass contribution of the element or group to the formula mas and multiply by 100;
Compound:
Ca₃(PO₄)₂
Formula mass = 3(40) + 2[31 + 4(16)]
= 120 + 2(95)
= 120 + 190
= 310
%C = 
x 100 = 38.7%
%P = 
x 100 = 41.3%
%O = 
x 200 = 20%
Answer:
The mass of copper sulfate in 25 cm³ of the copper sulfate solution is 0.45 g
Explanation:
The given parameters are;
The volume of the copper sulfate solution = 100 cm³
The mass of the copper sulfate in the solution = 1.8 g
Therefore, the mass of copper sulfate in 25 cm³ of the solution is given as follows;
The mass of copper sulfate in 100 cm³ of the solution = 1.8 g
The mass of copper sulfate in 1 cm³ of the solution = 1.8 g/100 = 0.018 g
Therefore;
The mass of copper sulfate in 25 × 1 = 25 cm³ of the solution, m = 25×0.018 g = 0.45 g
∴ The mass of copper sulfate in 25 cm³ of the solution, m = 0.45 g
Explanation:
a) Bohr model is perfect for atoms that have single electron and fortunately both Be3+ ion and H atom have one electron so, Bohr model can easily and accurately applied to predict the spectrum of Be3+ and H atom.
b) The energy of an atom in Bohr model is given by
the values of z for H atom and Be3+ ion are 1 and 4 respectively. Hence, energy of atoms would be different for both atoms. Hence, line spectra to be identical is not possible.
Answer:
146 g/mol → option b.
Explanation:
This is a problem about the freezing point depression. The formula for this colligative property is:
ΔT = Kf . m . i
We assume i = 1, so our compound is not electrolytic.
ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C
m = molality (mol of solute/kg of solvent)
We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.
We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1
1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg
0.19922 mol/kg = mol/ 0.250kg
mol = 0.19922 . 0.250kg → 0.0498 mol
molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol
