The answer is: the pressure inside a can of deodorant is 1.28 atm.
Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
p₁/T₁ = p₂/T₂.
p₁ = 1.0 atm.; initial pressure
T₁ = 15°C = 288.15 K; initial temperature.
T₂ = 95°C = 368.15 K, final temperature
p₂ = ?; final presure.
1.0 atm/288.15 K = p₂/368.15 K.
1.0 atm · 368.15 K = 288.15 K · p₂.
p₂ = 368.15 atm·K ÷ 288.15 K.
p₂ = 1.28 atm.
As the temperature goes up, the pressure also goes up and vice-versa.
Answer:
36.63 Torr
Explanation:
You need to use two expressions, one for pressure and the other with the relation of density and height of the column.
For the pressure:
P = h * d * g (1)
h is height.
d density
g gravity
The second expression put a relation between the densities and height of the column so:
d1/d2 = h1/h2 (2)
let 1 be the phthalate, and 2 the mercury.
Let's calculate first the relation of density:
d1/d2 = 13.53 / 1.046 = 12.93
Now with the first expression, we can calculate the pressure so:
P = hdg
We have two compounds so,
h1d1g = h2d2g ---> gravity cancels out
From here, we can solve for h2:
h2 = h1*(d1/d2)
replacing:
h2 = 459 / 12.53
h2 = 36.63 mm
1 mmHg is 1 torr, therefore the pressure of the gas in Torr would be 36.63 Torr
Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
