This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Cerebrum is the part of the brain that we use for imagination, thinking, judgement etc. It is also the largest part of the brain
Cerebellum is in charge of stuff like the spinal cord
and the brain stem is the part of the brain that helps us move. (if you watch the walking dead in one of the episodes it teaches this xD)
hope this helped (:
Answer:
C
Explanation:
Weather is what is happening at one point in time
Answer:
Total pressure in flask is 2.37 atm.
Explanation:
According to the Dalton's law of partial pressure, the total pressure of the flask would be the sum of partial pressure of the gases present in mixture.
P(total) = P1+ P2+P3+...Pn
n= number of gases present
Given data:
Pressure of argon gas = 0.72 atm
Pressure of oxygen gas = 1.65 atm
Total pressure in flask = ?
Solution:
P(total) = P (argon) + P (oxygen)
P(total) = 0.72 + 1.65
P(total) = 2.37 atm
Answer:
1.91 V
Explanation:
The correct cell reaction. equation is;
Br2(l) + H2(g) + 2OH-(aq) -> 2Br-(aq) + 2H2O(l)
Anode half reaction;
H2(g) + 2OH^-(aq) --------> 2H2O(l) +2e- -0.82 V
Cathode half reaction;
Br2(l) + 2e- -----> 2Br^-(aq). +1.09 V
Hence
E° cell= E°cathode - E°anode
E° cell= 1.09 - (-0.82)
E° cell= 1.91 V
