276 grams of carbon in 23.0 moles
Can you be more specific about what we are supposed to write
Answer:
1.64 L
Explanation:
P₁V₁ = P₂V₂
P₁ = 2.25 atm
V₁ = 0.75 L
P₂ = 1.03 atm
V₂ = ?
(2.25 atm)(0.75 L) = (1.03 atm)V₂
[(2.25 atm)(0.75 L)]/(1.03 atm) = V₂
V₂ = 1.64 L
Answer:
54.5%
Explanation:
The percentage composition of oxygen in C₆H₈O₆ can be obtained as follow:
Molar mass of C₆H₈O₆ = (12×6) + (8×1) + (16×6)
= 72 + 8 + 96
= 176 g/mol
Next, there are 6 oxygen atoms in C₆H₈O₆. Therefore the mass of oxygen in C₆H₈O₆ is:
Mass of Oxygen = 16 × 6 = 96 g
Finally, we shall determine the percentage composition of oxygen in C₆H₈O₆ as follow:
Percentage of oxygen =
Mass of Oxygen/mass of C₆H₈O₆ × 100
Percentage of oxygen = 96 / 176 × 100
Percentage of oxygen = 54.5%
Thus, the percentage composition of oxygen in C₆H₈O₆ is 54.5%.
100C is the total amount of heat added to the sample during interval CD
