11.2L/22.4L (STP value) x 1 mol of CH4 x 16.04 g of CH4 = 8.2 g
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answer:
S°m,298K = 85.184 J/Kmol
Explanation:
∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol
∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol
∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol
∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol
⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)
⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol
⇒ S°m,298K = 85.184 J/Kmol
[ H₃O⁺] = 10 ^ - pH
[ H₃O⁺ ] = 10 ^ - 7.30
[ H₃O⁺ ] = 5.011 x 10⁻⁸ M
hope this helps!
T<span>he charges themselves may not move fast, but the force upon them does. The electric field set up by the battery or generator propagates through the wires at the speed of light.
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