Answer:0.300M
Explanation:1) Data:
a) Initial solution
M = 1.50M
V = 50.0 ml = 0.050 l
b) Solvent added = 200 ml = 0.200 l
2) Formula:
Molarity: M = moles of solute / volume of solution is liters
3) Solution:
a) initial solution:
Clearing moles from the molarity formula: moles = M × V
moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol
b) final solution:
i) Volumen of solution = 0.050 l + 0.200l = 0.250l
ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer
Answer:
Explanation:
The polarity of the 3 compounds would be in the order of
Ferrocene < Acetylferrocene < Diacetylferrocene
Your TLC data has to also support this observation . This can be checked by measuring the values of Rf ( Retention factor = distance travelled by solute/solvent ) .The Rf values also has to follow this particular order: -
Ferrocene > acetylferrocene > diacetylferrocene
2) Hexane happens to be a non-polar solvent. The polarity of hexane can be increased if some polar solvents for example, ethyl and methylene chloride etc are added
Therefore, in the increasing order of solvents polarity, we have
Hexane < 1:1 mixture of hexane: methylene chloride < 9:1 mixture of methylene chloride:
3) Chromatographic techniques all have a stationary phase in addition to a mobile phase. In the case of column chromatography, the silica gel will be the stationary phase and the solvent that will be poured will be the mobile phase.
4) The TLC and column chromatography both happen to have the same stationary phase which is the silica gel. Also, the same solvent mixture is used in both the techniques. This makes the result of the 2 to be almost the same. The difference seen between them is that, TLC works against the gravity while on the other hand column chromatography works in the direction of the gravity.
5) The key feature in the IR spectra of the acetylferrocene that will be absent in the spectra of ferrocene is the presence of carbonyl stretching frequency at close to 1700 per cm(cm-1). This peak is easily differentiated between both acetyl ferrocene and ferrocene.
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.)
(46 sec) x √ ((253.80894 g I2/mol) / (44.0128 g N2O/mol)) = 110 sec
