Answer:
79.3%
Explanation:
Percent yield = 5.33g/6.72g x 100% = 79.3%
Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
Answer:
Option B
Explanation:
As Brønsted-Lowry theory states, acids are the ones that can donate protons.
When a proton is donated, it is released to become medium more acidic.
HCl is a strong acid.
HCl (l) + H₂O (l) → H₃O⁺ (aq) + Cl⁻(aq)
These always reffers to strong acid where the dissociation is 100% completed.
In a weak acid, dissociation is not 100% complete, that's why we have an equilibrium.
HA (l) + H₂O (l) ⇄ H₃O⁺ (aq) + A⁻(aq) Ka
Answer:
Calcium carbonate reacts with hydrochloric acid to form carbon dioxide gas. 2HCl (aq) + CaCO 3(s) CaCl 2 (aq) + CO 2(g) + H 2 O (l).
Answer:
4) Titration
Explanation:
Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.
