Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K
P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm
Answer:
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Answer:
Average atomic mass of carbon = 12.01 amu.
Explanation:
Given data:
Abundance of C¹² = 98.89%
Abundance of C¹³ = 1.11%
Atomic mass of C¹² = 12.000 amu
Atomic mass of C¹³ = 13.003 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100
Average atomic mass of carbon= 1186.68 + 14.43333 / 100
Average atomic mass of carbon = 1201.11333 / 100
Average atomic mass of carbon = 12.01 amu.
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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there are 6 moles of oxygen in Zn(No3)2
