Answer
<span>How does adding a non-volatile solute to a pure solvent affect the vapor pressure of the pure solvent?
</span>Answer The third option The solvent's vapor pressure will not be affected.
<span>To make a 2.0 M solution, how many moles of solute must be dissolved in 0.50 liters of solution?
Answer C. 1.0 mole solute.
</span>
Answer:
3 different elements are present in the compound.
Answer:
X(Cl-35) = 75.95% => Answer 'A'
Explanation:
34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance
X(Cl-35) + X(Cl-37) = 1 ⇒ X(Cl-37) = 1 - X(Cl-25)
34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45
34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45
Rearrange ...
36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45
2.0006·X(Cl-35) = 1.5195
X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance
⇒ % abundance = 75.95%
You would have to use the ideal gas law for this:
PV=nRT
Pressure, Volume, n=moles, R gas constant, Temperature in Kelvin
P=nRT/V
(1.8mol)(62.36)(309K)/43.0L = 805mm Hg
Answer:
6.63 M
Explanation:
NaCl(s) ---> Na^+(aq) + Cl^-(aq)
Given that [Na^+] = [Cl^-] = s
Where s= concentration of the both ions
Ksp = s^2
s= √Ksp
s= √43.9
s= 6.63 M
