Answer:
2266g
Explanation:
mass = no.of molecules /6.o23*1o(23) * molar mass
molar mass of co2= 44g /mol
1.5 .10^25/6.023 .10^23 =51.5 moles of co2
51.5 .44g/mol =2266 g
<span>Mg + O2 > MgO. In reactant side, 2 O atoms and 1 Mg are present. In product side, 1 Mg and O atoms are present. Put 2 in product side to balance O atoms and 2 at Mg in reactant side to balance Mg atoms. Therefore the balanced equation becomes, 2Mg + O2 ----> 2MgO. Hope it helps.</span>
Explanation:
b. What useful functions do oxidation numbers serve?
It is used to show oxidation and reduction (loss and gain of electrons)
b. How many molecules are in 1 mole of molecules?
1 mole = 6.022 * 10^23 molecules
c. What is the name given to the number of molecules in 1 mole?
Avogadro's Number of molecules
21. a. What is the molar mass of an element?
This is the mass of an element divided by the number of moles.
Molar mass = Mass / Number of moles
b. Write the molar mass rounded to two decimal places of carbon, neon, iron and uranium.
amu = Atomic Mass Unit
Carbon = 12.01 amu
Neon = 20.18 amu
Iron = 55.85 amu
Uranium = 238.03 amu
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =
Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality = 
Thus vant hoff factor for sodium chloride in X is 1.9
