Answer:
Can you please tell us what the following are?
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.
Protons/Electrons: 92
Neutrons: 140
***REMEMBER: There is always the same amount of protons and electrons. :)
Answer:
After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.
Explanation:
Answer:
The available amounts of each reactant
Explanation:
The limiting reactant is the reactant that will limit the equation.
For example, I have 2N2 and H2O. In this situation, H2O would be the limiting reactant since there is only one atom, while N2 has two atoms.
Think of it like you're making peanut butter and jelly. You have 50 cans of peanut butter and only 10 jelly. Which means you can only make sandwiches using 10 jars of jelly and 10 jars of peanut butter.