Response:
9.606 g
Clarification:
Step 1: Write the balanced combustion equation
C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the moles for 11.27 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
11.27 g × (1 mol/18.02 g) = 0.6254 mol
Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O
The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol
Step 4: Calculate the mass for 0.2085 moles of C₂H₆O
The molar mass of C₂H₆O is 46.07 g/mol.
0.2085 mol × 46.07 g/mol = 9.606 g
Answer:
Explanation:
Considering the reaction: 2X + 3Y = 3Z, combining 2.00 moles of X with 2.00 moles of Y results in the production of 1.75 moles of Z.
2 mol 2 mol 1.75 mol
2X + 3Y = 3Z
2 mol is required with 3 mol to yield 3 mol.
3 mol Z / 3 mol Y = 1 to 1
should yield 2 mol Z
1.75 / 2 = 87.5 % production yield
Ca3(PO4)2 is the correct formula.
Answer:
The original halide's formula is SrCl₂.
Explanation:
- The chemistry reaction's balanced equation is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.
- Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
- The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
- n = mass / molar mass, thus n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This identifies the atomic mass of Cl.
- Consequently, the original halide's formula is SrCl₂.
Answer:
(C) The average speed of molecules in ethane is the same as that of propanol.
Explanation:
In gas behavior, temperature is directly linked with speed. At a constant temperature, speed remains consistent. Also, we understand that ideal gases exhibit uniform behavior, irrespective of their type.
