Assuming 100% dissociation, calculate the freezing point and boiling point of a 2.5 m sncl4 (aq)

The freezing point of the solution is 249.75 K and boiling point of the solution is 379.4 K.

Explanation :

Molality of = 2.5

a) Depression in freezing point:

i = Van'T Hoff factor =  5 (for 100% dissociation)

= freezing point constant= 1.86Kkg/mol

The freezing point of the solution is .

b) Elevation in boiling point:

i = Van'T Hoff factor =  5 (for 100% dissociation)

=boiling point constant= 0.512 Kkg/mol

The boiling point of the solution is .

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2