168.96 g of carbon dioxide (CO₂)
Explanation:
The chemical reaction representing the combustion of acetylene:
2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)
number of moles = mass / molecular weight
number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)
then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)
X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)
mass = number of moles × molecular weight
mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g
Learn more about:
combustion of hydrocarbons
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Answer:
212.5 mL
both the original and the diluted solution have 0.765 moles of KCl
Explanation:
c1V1 = c2V2
V2 = c1V1/c2 = (1.8 M×425 mL)/1.2 M = 637.5 mL
(637.5 - 425) mL = 212.5 mL
n = (1.8 mol/L)(0.425 L) = 0.765 moles of KCl
since it's a dilution, the diluted solution has the same number of moles as the original solution, 0.765 moles of KCl
Answer:
100 degree celcius, because it is the melting point of ice ob
First, we calculate the mass of the sample:
mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams
Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.
Zinc mass = 0.37 x 954
Zinc mass = 352.98 g
Zinc moles = 352.98 / 65
Zinc moles = 5.43
Zinc atoms = 5.43 x 6.02 x 10²³
Zinc atoms = 3.27 x 10²⁴
Copper mass = 0.63 x 954
Copper mass = 601.02 g
Copper moles = 601.02 / 64
Copper moles = 9.39
Copper atoms = 9.39 x 6.02 x 10²³
Copper atoms = 5.56 x 10²⁴
To determine the mass of oxygen per gram of sulfur for sulfur dioxide, we simply obtain the ratio of the mass of oxygen and the mass of sulfur produced from the decomposition of sulfur dioxide. All other values given in the problem statement above are just to confuse us that the question is a difficult one. We do as follows:
mass of oxygen per gram sulfur = 3.45 g / 3.46 g
mass of oxygen per gram sulfur = 0.9971 g O2 / g S
