Answer:
continental polar air masses
Explanation:
forms over Canada, move down from North Pole, bring bitterly cold and dry air to the northern United States during the winter.
Answer: 10
Explanation:
The detailed solution is contained in the image attached. The molar mass of hydrated and anhydrous salts are obtained and the number of moles of hydrated and hydrated salts are equated. The masses of hydrated and anhydrous salts are gives in the question and are simply substituted accordingly. This can now be used to obtain the number of molecules of water of crystallization as required in the question.
<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

Or,

We are given:
Mass of iron (III) chloride = 1.01 g
Molar mass of iron (III) chloride = 162.2 g/mol
Volume of the solution = 10 mL
Putting values in above equation, we get:

Hence, the molarity of Iron (III) chloride is 0.622 M.
Answer:
m = E × Q
And Q = I × t
m = E × I × t
Where m = mass in grams
Q = quantity of electricity in coulomb
I = current in ampere
t = time in seconds
E = electrochemical equivalent of the substance
Explanation: