Answer:
4.67M
Explanation:
The concentration of methanol (CH3OH) can be calculated using the following:
Molarity (M) = number of moles(n)/volume(v)
However, mole is not given. It can be obtained by using:
Mole = mass / molar mass
Where; mass = 34.4g
Molar mass (MM) of CH3OH is:
= 12 + 1(3) + 16 + 1
= 12 + 3 + 17
= 32g/mol
mole = 34.4/32
mole = 1.075mol
The volume needs to be converted to L by dividing by 1000
230mL = 230/1000
= 0.230L
Molarity = mol/volume
Molarity = 1.075/0.230
Molarity = 4.6739
Molarity = 4.67M
The concentration of CH3OH in solution is 4.67M
Answer:
5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)
Explanation:
mole conversions:
grams to moles => divide by formula wt.
moles to grams => multiply by formula wt.
gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)
This problem:
mass to moles => divide by formula wt.
mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)
Help 2 what the answer for this?
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
