Answer:
Yes, it is possible.
Explanation:
A diprotic acid is an acid that can release two protons. That's why it is called diprotic.
Monoprotic → Release one proton, for example Formic acid HCOOH
Triprotic → Releases three protons, for example H₃PO₄
Polyprotic → Release many protons, for example EDTA
it is a weak acid.
In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
HA⁻ + H₂O ⇄ OH⁻ + H₂A Kb₂
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.
Answer:
5 pounds 2 ounces.
Explanation:
1 pound = 16 ounces
So just add the 2 ounces to the pounds.
Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
