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.
The question is incorrect, the correct question is;
Carbon-14, which is present in all living tissue, radioactively decays via a first-order process. A one-gram sample of wood taken from a living tree gives a rate for carbon-14 decay of 13.6 counts per minute. If the half-life for carbon-14 is 5720 years, how old is a wood sample that gives a rate for carbon-14 decay of 11.9 counts per minute?
Answer:
C. 1.1 x 103 yr
Explanation:
A = count rate of the wood sample
Ao= count rate of a living tissue
t1/2= half life of C-14
t = time taken
From;
0.693/t1/2 = 2.303/t log Ao/A
0.693/5720 = 2.303/t log (13.6/11.9)
1.2 * 10^-4 = 2.303/t * 0.05799
1.2 * 10^-4 = 0.1336/t
t = 0.1336/1.2 * 10^-4
t = 1.1 x 103 yr
Answer:
C₈H₈O₃
Explanation:
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of C:H:O.
Assume 100 g of the compound.
1. Calculate the mass of each element.
Then we have 63.15 g C, 5.30 g H, and 31.55 g O.
2. Calculate the moles of each element
![\text{Moles of C} = \text{63.15 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{5.258 mol C}\\\\\text{Moles of H} = \text{5.30 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{5.258 mol H}\\\\\text{Moles of O} = \text{31.55 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{1.972 mol O}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20C%7D%20%3D%20%5Ctext%7B63.15%20g%20C%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20C%7D%7D%7B%5Ctext%7B12.01%20g%20C%7D%7D%20%3D%20%5Ctext%7B5.258%20mol%20C%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20H%7D%20%3D%20%5Ctext%7B5.30%20g%20H%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20H%7D%7D%7B%5Ctext%7B1.008%20g%20H%20%7D%7D%20%3D%20%5Ctext%7B5.258%20mol%20H%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20O%7D%20%3D%20%5Ctext%7B31.55%20g%20O%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20O%7D%7D%7B%5Ctext%7B16.00%20g%20O%20%7D%7D%20%3D%20%5Ctext%7B1.972%20mol%20O%7D)
3. Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
C:H:O = 5.258:5.258:1.972 = 2.667:2.666:1
4. Multiply by a number to make each ratio close to an integer
Multiply the ratios by three.
2.667:2.666:1 = 8.000:8.000:3 ≈ 8:8:3
5. Write the empirical formula
EF = C₈H₈O₃
Answer:
25.0%
A solution of benzene (C6H6) and toluene (C7H8) is 25.0% benzene by mass. At 25 °C the vapor pressures of pure benzene and pure toluene are 94.2 torr and 28.4 torr, respectively.