Answer:
x-10 should be the answer
Answer:
i don't think you can, but if you do know let me know
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
Answer:
D) 1.61 times faster
Explanation:
= √(3)RTM
R constant= 0.08206
T=constant, so in this problem we dont need a value for it
M=17.031 g/mol
√(3)(0.08206)(17.031)= 2.047
= √(3)RTM
R constant= 0.08206
T=constant, so in this problem we dont need a value for it
M= 44.01 g/mol
√(3)(0.08206)(44.01)= 3.29
Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.
So instead of doing (2.047)/(3.29) = 6.22
we have to flip the values to get (3.29)/(2.047)= 1.61
Answer:
higher
Explanation:
If it contains a impurity that is insoluble it won't dissolve completely, the solution would be smaller than it is supposed to be, when compared to a compound without such insoluble impurity. Molecular weight determination won't be accurate because the molecular weight obtained will be higher as a result of the fact that the mass of the solute would include the actual solute that is changing the temperature and the excess mass of the impurity.
