Which term is same for one mole of oxygen and one mole of water
1 answer:
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Explanation:
First, calculate the moles of using ideal gas equation as follows.
PV = nRT
or, n =
= (as 1 bar = 1 atm (approx))
= 0.183 mol
As, Density =
Hence, mass of water will be as follows.
Density =
0.998 g/ml =
mass = 3.25 g
Similarly, calculate the moles of water as follows.
No. of moles =
=
= 0.180 mol
Moles of hydrogen = = 0.36 mol
Now, mass of carbon will be as follows.
No. of moles =
0.183 mol =
= 2.19 g
Therefore, mass of oxygen will be as follows.
Mass of O = mass of sample - (mass of C + mass of H)
= 3.50 g - (2.19 g + 0.36 g)
= 0.95 g
Therefore, moles of oxygen will be as follows.
No. of moles =
=
= 0.059 mol
Now, diving number of moles of each element of the compound by smallest no. of moles as follows.
C H O
No. of moles: 0.183 0.36 0.059
On dividing: 3.1 6.1 1
Therefore, empirical formula of the given compound is .
Thus, we can conclude that empirical formula of the given compound is .
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).