Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q
To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.
The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.
To learn more about crystal field stabilization energy visit:brainly.com/question/29389010
#SPJ4
One of the atoms involved must be a metal and the other a nonmetal
Answer:
See detailed explanation.
Explanation:
Hello there!
In this case, according to the given description, it turns out possible for us to infer that the second fractionating column on top of the first one will favor the light product, in this case hexane as it has the lowest boiling point and molar mass; in such a way, we can tell the following:
a) The separation between hexane and heptane will be increased as a purer hexane-rich product will be obtained on the top of the second column.
b) Will be increased as well, because the second column will remove more heptane.
c) Also, more pure heptane will be obtained on the bottom of the two columns, yet the most favored yield will be that of hexane.
All of the aforementioned is possible due to the fact that the second column will remove the amount of heptane that could not be removed on the top of the first column by taking the vapor-liquid equilibrium further from the first column's maximum separation, which is known as distillation sequences.
Regards!
