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:
1.40*10⁻² M
Explanation:
We have the solubility formula
Solubility,
S = KH*P
where
KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm
P = atmospheric pressure = 0.400 atm
Hence, we have
S = KH*P
= (3.50*10⁻² mol/L.atm)*(0.400 atm)
= 1.40*10⁻² mol/L
But 1 mol/L = 1 M,
Hence, the answer (1.40*10⁻² mol/L
) is equivalent to
= 1.40*10⁻² M
Answer: 1.00×10^26=1000×10^23
1000×10^23/6.022×10^23=166.05×10^23 moles of water.
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Explanation: