Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:
C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)
A 0.035 M solution of ephedrine has a pH of 11.33.
a) What are the equilibrium concentrations of C10H15ON, C10H15ONH<span>+, and OH-</span>?
b) Calculate <span>Kb</span> for ephedrine.
c(C₁₀H₁₅NO) = 0,035 M.<span>
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH</span>⁻] =
10∧(-2,67) = 0,00213 M.<span>
[OH</span>⁻] =
[C₁₀H₁₅NOH⁺] = 0,00213 M.<span>
[</span>C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.<span>
Kb = [OH</span>⁻] ·
[C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].<span>
Kb = (</span>0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
Endothermic<span> Reaction??? </span>
Answer:
The first row of elements fits in period <u>6</u>, after the element <u>lanthanum (La)</u>. The second row of elements fits in period <u>7</u>, after the element <u>actinium (Ac). </u>
I hope this helps!