The complete balanced chemical equation for this is:
<span>3KOH + H3PO4
--> K3PO4 + 3H2O</span>
First we calculate the number of moles of H3PO4:
moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol
From stoichiometry, 3 moles of KOH is required for every
mole of H3PO4, therefore:
moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole
H3PO4) = 0.0468 mol
Calculating for volume given molarity of 0.350 M KOH:
Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7
mL
Answer:
<span>133.7 mL KOH</span>
Ionization energy is the measure of the extend to which the nucleus attracts the outermost electron
if ionization energy us high than force of attraction Is high so it is not easy to remove and vice versa .
hope you understand.....
Answer:
mass PbI₂ formed = 1383 grams
Explanation:
Pb(NO₃)₂ + 2NaI => 2NaNO₃ + PbI₂(s)
6 mol NaI => 1/2(6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
