We have been given the condition that carbon makes up 35%
of the mass of the substance and the rest is made up of oxygen. With this, it
can be concluded that 65% of the substance is made up of oxygen. If we let x be
the mass of oxygen in the substance, the operation that would best represent
the scenario is,
<span> x = (0.65)(5.5 g)</span>
<span> <em> </em><span><em>x =
3.575 g</em></span></span>
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).
Answer:
Option d. 2615.0g
Explanation:
Let M1, M2, and M3 represent the masses of the three different samples
M1 = 0.1568934 kg = 156.8934g
M2 = 1.215mg = 1.215x10^-3 = 0.001215g
M3 = 2458.1g
Total Mass = M1 + M2 + M3
Total Mass = 156.8934 + 0.001215 + 2458.1
Total Mass = 2614.994651g
Total Mass = 2615.0g
Answer: 1.348 ×10^23 atoms
Explanation:
Given that volume = 1.00L
At standard condition, the volume of a gas is 22.4L/mol (at S.T.P)
Volume = mole /volume at STP
1 = mole/22.4
Mole= 22.4mol.
Also
Mole = number of atoms /Avogradro constant
Where avogrado's constant = 6.02×10²³
22.4 = number of atoms/6.02×10²³
Number of atoms = 1.348×10^25atoms
