Answer:
4.1 atm = 3,116 mmHg = 415.4 kPa
Explanation:
According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:
P₁ x V₁= P₂ x V₂
Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.
From the problem, we have:
V₁= 50.0 L
V₂= 68.0 L
P₂= 3.0 atm
Thus, we calculate the initial pressure as follows:
P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm
To transform to mmHg, we know that 1 atm= 760 mmHg:
4.1 atm x 760 mmHg/1 atm = 3,116 mmHg
To transform to kPa we use: 1 atm= 101.325 kPa
4.1 atm x 101.325 kPa = 415.4 kPa
Erica is likely to use reverse osmosis process.it is the method used in the industry to treat water in oil and gas industry .or ultra filtration method is also employed
You need to use the ideal gas law (PV=nRT) and solve for n. ((3.50atm•10.0L)/(0.0821(L•atm/mol•K)•304K) = n = 1.40 moles. 1 mole of Cl2 = 70.9 gm/mole. The mass would be 99.43 gm
To assume the empirical formula of a compound, you want the ratio of the moles of every element, and you discover that by means of the percent’s of the element as the element's mass.
As an instance, if a compound is 16% Carbon and 84% sulfur, you can round about that if you had a 100 gram sample of the compound, it would contain 16 grams of carbon & 84 grams of sulfur.
To look for the moles of carbon in that sample, you would distribute the mass by the atomic mass of carbon, so 16/12 = 1.3 moles. You do the similar calculation with the other elements. For Sulfur, you divide 84g by the atomic mass of sulfur, so 84/32 = 2.6moles of sulfur. You endure in this same way if there is more than 2 elements.
Lastly you find the ratio of the moles of every element. The unassuming way to do this is to look the element with the smallest number of moles and split the other moles by that number. In the above example 2.6 moles of Sulfur divided by 1.3 moles of Carbon equals 2. (Which is a 2:1 ratio) Therefore there is twice as numerous sulfurs as carbons in this compound, and the empirical formula is CS2.
