Answer:
3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
Answer:
Sun and the stars, nothing humans don't use fusion at all
Explanation:
Answer: in a straight path!
Explanation: hope this helps
In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.
Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.
Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M
The concentration of HBr is 0.127 M.