Answer:
Explanation:
It depends on how this is done. If you raise the pressure, the nitrogen will disappear (liquify) and all that will be left will be the 21 % oxygen and the 1% argon.
The process is very complicated because the boiling point of nitrogen keeps on changing. The boiling point is unstable.
Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
Answer:
See explanation
Explanation:
In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.
See figure 1
I hope it helps!
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
