Molarity
a. Na2CO3
Molar mass = 106 g/mol
n= 109 g / (106 g/mol) = 1.028 mol
b. 6.00L
c. M=n/V
M = 1.028 mol / 6.00 L
M= 0.1713 mol / L
Answer: The molarity of the sodium carbonate solution is 0.1713 mol/L
Answer:
B.Carbohydrates and D. water
Answer:- 23.0 mg
Solution:- Radioactive decay obeys first order kinetics and the first order kinetics equation is:
where, 
is the initial amount of radioactive substance and N is it's amount after time t. k is the decay constant.
From given information, Original amount, of the radioactive substance is 184 mg and we are asked to calculate the amount N after 15 days. It means, t = 15 days
Half life is given as 5 days. From the half life, we could calculate the decay constant k using the equation:
where, 
is the symbol for half life. let's plug in the value of half like to calculate k:
Let's plug in the values in the first order kinetics equation and solve it for N:
lnN = 3.136
N = 23.0 mg
So, 23.0 mg of Bi-210 would be remaining after 15 days.
Answer:
4.12 moles
Explanation:
We can solve this problem with the Ideal Gases Law.
P . V = n . R . T
In our first case we have:
P = 2.3 atm
V = 32.8 L
n = 2.98 moles
T → 35°C + 273 = 308K
Let's replace data for the second case:
2.3 atm . 45.3L = n . 0.082 . 308K
n = (2.3 atm . 45.3L) / (0.082 L.atm/mol.K . 308K)
n = 4.12 moles
