Answer:
0.3267 M
Explanation:
To solve this problem, first we calculate how many moles of Mn(ClO₄)₂ are contained in 23.640 g of Mn(ClO₄)₂·6H₂O.
Keep in mind that the crystals of Mn(ClO₄)₂ are hydrated, and <em>we need to consider those six water molecules when calculating the molar mass of the crystals</em>.
Molar mass of Mn(ClO₄)₂·6H₂O = 54.94 + (35.45+16*4)*2 + 6*18 = 361.84 g/mol
Now we <u>proceed to calculate</u>:
- 23.640 g Mn(ClO₄)₂·6H₂O ÷ 361.84 g/mol = 0.0653 mol Mn(ClO₄)₂·6H₂O = mol Mn(ClO₄)₂
Now we divide the moles by the volume, to <u>calculate molarity</u>:
- 200 mL⇒ 200/1000 = 0.200 L
- 0.0653 mol Mn(ClO₄)₂ / 0.200 L = 0.3267 M
The closest shell (n = 1) can contain a maximum of 2 electrons.
3) CH₃-COOH + NH₃ → CH₃-COO⁻NH₄⁺
4) 2 FeCl₃ + 3 Ag₂SO₃ → Fe₂(SO₃)₃ + 6 AgCl
5) 2 Al + 3 NiCl₂ → 2 AlCl₃ + 3 Ni
6) 4 LiCl + Pb(NO₂)₄ → 4 LiNO₂ + PbCl₄
7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O
8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃
9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄
the correct answer is.. ill tell you when i search it up gimme 2 seconds
