<span>When
a solute is added to a solvent, some properties are affected and these set of
properties are called colligative properties. The properties depend on the amount of solute
dissolved in a solvent. For this we use the boiling point elevation of the solution. We do as follows:
</span>ΔT(boiling
point) = 103.06 °C - 100.0 °C= 3.06 °C<span>
ΔT(boiling point) = (Kb)m
m = </span>ΔT(boiling point) / Kb<span>
m = 3.06 °C / 0.512 °C kg / mol
m = 5.98 mol / kg
</span>
C. The Seasons change during the year.
Would be the correct answer.
Bubbling if you put it in a liquid
CH3OCH3 and C2H5OH are isomers. They have same molecular formula. But with another formation.
Answer:
A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
(4x^2)x
Kc= -----------
(A-2x)^2
PV=nRT
n/v = P/RT = .745/(0.0821)(334+273) = .01495
To Find the initial molarity of NO2
(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L
Thus:
46(A-2x) + 2x(30) + 32x = .515 g/L
46A-92x+60x+32x = .515
46A=.515
A=.01120 M
Using the total molarity found
(A-2x)+2x+x = .01495 M
A+x=.01495
Plug in A found into the above equation:
.01120+x = .01495
x=.00375
Now Plug A and x into the original Equilibrium Constant Expression:
(4x^2)x
Kc= -----------
(A-2x)^2
Kc = 0.000014
Explanation:
