Answer:
- Dipole interactions
- London dispersion forces
- Hydrogen bonds
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Development of carbonation
Answer: the percent composition of carbon in heptane is 83.9%
Explanation:
<u>1) Atomic masses of the atoms:</u>
<u>2) Molar mass of heptane:</u>
- C₇H₁₆: 7 × 12.01 g/mol + 16×1.008 g/mol = 100.2 g/mol
<u>3) Mass of carbon in one mole of heptane:</u>
- C₇: 7 × 12.01 g/mol = 84.07 g/mol
<u>3) Percent composition of carbon:</u>
- % = (mass in grams of C) / (mass in grams of C₇H₁₆) × 100 =
= (84.07 g/ 100.2 g) × 100 = 83.9% ← answer
Answer:
D. 1.48atm
Explanation:
Van der waals equation is given as:
(P +an²/v²) (v - nb) = nRT
Where;
P = pressure (atm)
V = volume (L)
R = gas constant (0.0821 Latm/molK)
a and b = gas constant specific to each gas
T = temperature (K)
n = number of moles
According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol
Hence;
(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15
(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638
(P + 0.0291) (22.316) = 33.638
22.316P + 0.649 = 33.638
22.316P = 33.638 - 0.649
22.316P = 32.989
P = 32.989/22.316
P = 1.478
P = 1.48atm