Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.
Refer to the attached document for the solution.
Every unicellular organism prospers by executing metabolic activities.
Metabolic activities encompass the set of chemical reactions essential for sustaining life.
Explanation:
Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.
These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.
Organisms acquire the energy necessary for these activities through food consumption.
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I believe you mean KO2 reacting with H2O. The reaction is 4KO2+2H2O->4KOH +3O2. The mole ratio O2:KO2 is 3:4. Thus moles of O2 produced = 0.500/4*3 = 0.375 mol.
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.
