<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer:
Number of moles of methane form = 2.3 mol
Explanation:
Given data:
Number of moles of Hydrogen = 4.6 mol
Number of moles of methane form = ?
Solution:
Chemical equation:
C + 2H₂ → CH₄
Now we will compare the moles of methane with hydrogen from balance chemical equation.
H₂ : CH₄
2 : 1
4.6 : 1/2×4.6 = 2.3 mol
Form 3.6 moles of hydrogen 2.3 moles of methane can be formed.
Answer:
In short, because we are tearing up the oxygen factories to make way for carbon dioxide emitters. (Doesn't make a lot of sense, read the explanation)
Explanation:
So, 1000 years ago, we had a lot more trees, didn't have engines or cars or factories or anything, really that released carbon dioxide into the air and we had a lot more trees and since the invention of cars, engines, carbon dioxide-emitting tools and factories and all the other things that emit "Greenhouse gases" and in doing that, cleared more trees to make room for factories and roads and that has drastically changed the outlook of the carbon cycle.
Answer:
in the periodic table we can see that the ions of Cl is greater than the ions in Na
