The answer is 492.8 g
1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample.
1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole
6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n
n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol
2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:
Ar(C) = 12 g/mol
Ar(H) = 1 g/mol
Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol
3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:
m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
Explanation:
The transuranium elements are produced by the capture of neutrons
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Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
