First, you need to find:
One mole of
is equivalent to how many grams?
Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of <span>Chlorine Cl = 35.45 g (See in group 17)
</span>
As there are two atoms of Chlorine present in
, therefore, the atomic mass of
would be:
Atomic mass of
= Atomic mass of Ca + 2 * Atomic mass of Cl
Atomic mass of
= 40.078 + 2 * 35.45 = 110.978 g
Now,
110.978 g of
= 1 mole.
75.9 g of
=
= 0.6839 moles.
Hence,
The total number of moles in 75.9g of
= 0.6839 moles
According to <span>Avogadro's number,
1 mole = 1 * </span>
molecules
0.6839 moles = 0.6839 *
molecules =
molecules
Ans: Number of molecules in 75.9g of = molecules
-i
Answer: The molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Explanation:
Molarity is the number of moles of a substance present in liter of a solution.
And, moles is the mass of a substance divided by its molar mass.
(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.
Now, molarity of ethanol solution is as follows.
(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.
Now, molarity of sucrose solution is as follows.
(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.
Now, molarity of sodium chloride solution is as follows.
Thus, we can conclude that the molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Answer:
temperature and number of molecules of gas
Explanation:
Boyle's law is one of the gas laws, which states that the volume of a gas is inversely proportional to the pressure at a constant temperature. The Boyle's law equation is given as follows:
P ∝ (1/V)
P = K/V
PV = K
Based on these, the temperature of an ideal gas and the number of molecules in the gas are kept constant in Boyle's law.
Answer:
Exothermic
Explanation:
Exothermic reactions release heat, endothermic absorb heat, and precipitate is a solid formed from a solution
Because atoms weigh much less than a gram.