Answer:

Explanation:
Hello,
In this case, we write the reaction again:

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

Best regards.
Molarity = Moles/Liter
Use the molecular atomic mass of NaCl to convert from grams to moles.
Molecular mass of NaCl is the sum of its atomic masses. Look at the periodic table to find these. Na is 23 g/mol and Cl is 35.5 g/mol ,
so NaCl = 23 + 35.5 = 58.5 g/mol
multiply to cancel out grams
76 g NaCl * (1mol / 58.5 g NaCl) = 1.3 mol NaCl
over 1 Liter is just 1.3 M NaCl
Hope this helps!
Answer:
Explanation:
By definition, <em>half neutralization</em> is the point at which half of the acid has been neutralized.
The neutralization reaction that you are studying is the acid-base reaction:
- HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)
Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:
- Half molarity = M / 2 = 0.2 M / 2 = 0.1 M
So, the answer is the first choice: a. 0.1 M.
Nonessential amino acids can be made by the body, while essential amino acids cannot be made by the body so you must get them from your diet.
Nonessential acids : glutamine, proline, glycine etc
Essential acids: lysine, leucine etc.
Use equation
number of moles= sample mass/molar mass