You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 20 g
m (final mass after time T) = 5 g
x (number of periods elapsed) = ?
P (Half-life) = ? (in minutes)
T (Elapsed time for sample reduction) = 8 minutes
Let's find the number of periods elapsed (x), let us see:






Now, let's find the half-life (P) of the radioactive sample, let's see:





I Hope this helps, greetings ... DexteR! =)
Answer:
0.08 g
Explanation:
100.0 mL = 0.10 L
Multiply the volume by the molarity to find moles.
0.10 L × 0.20 M = 0.002 mol
Convert moles to grams.
0.002 mol × 40 g/mol = 0.08 g