Reaction of calcium with water
Calcium reacts slowly with water. This is in contrast with magnesium, immediately above calcium in the periodic table, which is virtually unreactive with cold water. The reaction forms calcium hydroxide, Ca(OH)2 and hydrogen gas (H2).
Answer: It is true that Your friend should release the
first.
Explanation:
The velocity of particles of a gas is inversely proportional to the mass of gas. This means that more is the mass of gas less will be its velocity.
Or, more will be the mass of gas more slowly it will move from one place to another.
The molar mass of chlorine gas is more than the molar mass of hydrogen gas. Therefore, chlorine gas will move slowly.
So, your friend should release the
gas first and then according to the length of room you should release the
gas.
Thus, we can conclude that it is true that Your friend should release the
first.
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
Answer:
146.85 g/mol
Explanation:
PV=nRT
n=mass/molar mass
covert from mmhg to atm = 0.184 atm
convert from ml to L= 0.108 L
convert from degree C to K= 456.15 K
convert from mg to g= 0.07796g
then rearrange the formula:
n=PV/RT
=(0.184)(0.108)/(0.08206)(456.15)
n= 5.308*10^(-4)
rearrange the n formula interms of molar mass:
Molar mass= mass/n
=0.07796/(5.308*10^-4)
molar mass= 146.85g/mol
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