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
<span>Ka is an equilibrium constant for the partial ionization of "weak" acids in water.</span>
Newton's first law is the answer.
Answer:
YIKES. a bit late. Answers include 1, 2, 3
Explanation:
N2 + 3H2 ----> 2NH3
<span>you can see 3 moles H2 reacts to form 2 moles NH3 </span>
<span>Therefore moles NH3 = 2 / 3 x moles H2 </span>
<span>= 2/3 x 12.0 mol </span>
<span>= 8.00 mol NH3 hope this help</span>
