Answer:
2 Answers. The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by tM in the diagram.
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:
K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..
Then you can find the ration between K's at two different temperatures as:
K1 = A * e ^ (-Ea / RT1)
K2 = A* e ^(-Ea / RT2)
=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }
=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }
=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }
=> K1 / K2 = e ^ (2.0134494) ≈ 7.5
Answer: 7.5
Answer:
23.34 %.
Explanation:
- The percentage of water must be calculated as a mass percent.
- We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
- <em>Atomic masses of the elements:</em>
Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.
- <em>Mass of the formula unit:</em>
Cd(NO₃)₂•4H₂O
mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.
- <em> Mass of water in the formula unit:</em>
<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.
- <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
