The pressure of the gas is expected to increase in accordance to Boyle's law.
<h3>What is Boyle's law?</h3>
Boyle's law states that, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.
By implication, when the piston is lowered and the volume of the gas is decreased, the pressure of the gas is expected to increase in accordance to Boyle's law.
Learn more about Boyle's law: brainly.com/question/1437490
The power used by Alex to drag the log across the yard is determined as 2,656 W.
<h3>Mass of the log</h3>
The mass of the log is calculated as follows;
W = mg
m = W/g
m = (400)/9.8
m = 40.82 kg
<h3>Velocity of the log</h3>
K.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 900)/(40.82)
v² = 44.096
v = 6.64 m/s
<h3>Power used by Alex</h3>
P = Fv
P = 400 x 6.64
P = 2,656 W
Learn more about power here: brainly.com/question/13881533
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Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
Rotting, and ageing. The banana peel ages faster in open air, and therefor begins to rot. Brow peels don't mean that the banana is rotted, however, it is a sign that it may begin to rot soon.
16 I think but I could be wrong lol
