The atomic mass of a certain element is summation of the product of the decimal equivalent of the percentage abundance and the given atomic mass of each of the isotope. If we let x be the percentage abundance of the 86 amu-isotope then, the second one is 1-x such that,
x(86) + (1 - x)(90) = 87.08
The value of x from the equation is 0.73. This value is already greater than 0.5. Thus, the isotope with greatest abundance is that which is 86 amu.
Answer:
Na₂CO₃ · 10H₂O
Explanation:
The formula for sodium carbonate hydrate is:
Na₂CO₃ · xH₂O
The unknown "x" is the number of water molecules contained in the hydrate.
To find "x" we have to use the hydrogen percentage in the sample, 7.05 % H.
First we calculate the molecular weight of Na₂CO₃ · xH₂O:
molecular weight of Na₂CO₃ · xH₂O = 23 × 2 + 12 + 16 × 3 + 18x
molecular weight of Na₂CO₃ · xH₂O = 106 + 18x g/mole
Now we devise the fallowing reasoning tanking in account 1 mole of Na₂CO₃ · xH₂O:
if in 106 + 18x grams of Na₂CO₃ · xH₂O we have 2x grams of hydrogen
then in 100 grams of Na₂CO₃ · xH₂O we have 7.05 grams of hydrogen
106 + 18x = (100 × 2x) / 7.05
106 + 18x = 28.4x
106 = 28.4x - 18x
106 = 10.4x
x = 106 / 10.4
x = 10.2 ≈ 10
The formula for the washing soda is Na₂CO₃ · 10H₂O.
C2H6O + O2 ---> C2H4O2 + H2O
using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g
60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically
But as the yield is only 7.5%
the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g
= 7.156 kg to nearest gram. Answer
Answer:
CO
Explanation:
From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.
t1/t2=√m1/m2
Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g
1.25/1=√44/m2
Therefore m2=28g CO
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
