Answer: The empirical formula is 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mas of H = 1.8 g
Mass of S = 56.1 g
Mass of O = 42.1 g
Step 1 : convert given masses into moles.
Moles of H =
Mass of S =
Moles of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For H = 
For S =
For O =
Converting to whole number ratios
The ratio of H: S: O= 2: 2: 3
Hence the empirical formula is
Answer:
356.484 g.
Explanation:
- Molarity (M) is defined as the no. of moles of solute (NaCl) dissolved in a 1.0 L of the solvent.
<em>M = (no. of moles of NaCl)/(volume of the solution (L))</em>
M = 6.10 M, volume of the solution = 1.0 L.
∴ No. of moles of NaCl = (M)(volume of the solution (L)) = (6.10 M)(1.0 L) = 6.10 mol.
<em>∵ no. of moles of NaCl = (mass of NaCl)/(molar mass of NaCl)</em>
∴ Mass of NaCl = (no. of moles of NaCl)(molar mass of NaCl) = (6.10 mol)(58.44 g/mol) = 356.484 g.
Answer:
- 2977 kJ
Explanation:
Let's consider the following reaction.
3 CH₄(g) + 4 O₃(g) → 3 CO₂(g) + 6 H₂O(g)
We can find the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = [3 mol × ΔH°f(CO₂(g)) + 6 mol × ΔH°f(H₂O(g))] - [3 mol × ΔH°f(CH₄(g)) + 4 mol × ΔH°f(O₃(g))]
ΔH°r = [3 mol × (-393.5 kJ/mol) + 6 mol × (-241.8 kJ/mol)] - [3 mol × (-74.87 kJ/mol) + 4 mol × (142.7 kJ/mol)]
ΔH°r = - 2977 kJ
