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.
Convert the child weight (37.3 pounds) to kilograms
37.3 lb x 0.453 kg /1lb = "A kg"
multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him
A kg * 9.00 mg/1kg = "B mg"
calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL
B mg * 1mL/ 60.0 mg = C mL <span>oxcarbazepine</span>
Most atoms have more neutrons than protons which accounts for basically half of the nucleuses mass.
v=fw (Assume for this example w is wavelength). w=v/f. w=100/1000= 0.1 m. The wavelength is 0.1 meters