Well one mole of stuff, any stuff, including carbon dioxide, specifies
6.022
×
10
23
individual items of that stuff.
Explanation:
And thus we work out the quotient:
7.2
×
10
25
⋅
carbon dioxide molecules
6.022
×
10
23
⋅
carbon dioxide molecules
⋅
m
o
l
−
1
≅
120
⋅
m
o
l
carbon dioxide
.
This is dimensionally consistent, because we get an answer with units
1
m
o
l
−
1
=
1
1
mol
=
m
o
l
as required.
Answer:
13.93849 millimeters/second
Explanation:
The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m
Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.
<span>1.61 × 1023 Multiply by 26.8 to get the answer.161.33 x 10 ^23 </span>
