second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
Answer:
Common combustion reactions break the bonds of hydrocarbon molecules,
Explanation:
the resulting water and carbon dioxide bonds always release more energy than was used to break the original hydrocarbon bonds. That's why burning materials mainly made up of hydrocarbons produces energy and is exothermic.
Answer: None of the above
Explanation: It is not possible for any of these animals to undergo metamorphosis.