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 is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) <span>is a temporary attractive </span>force between molecules.
i dont have paper but heres where they should go
Answer: D. Mutation in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.
Explanation: It was not likely to be that the coding sequences are replicated more often. The only possible explanation is that the mutations in coding is more likely to be deleterious to the organism than mutations because it is in a non coding sequence.
HNO3+KOH = H2O+KNO3 . When nitric acid react with pottasuim hydroxide, the reaction will produce water (H20) and pottasuim trioxonitrate
