Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
Endothermic reactions. These are reactionsthat take in energy from the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surroundings to get colder
Answer:
C. It does not participate in a decay series.
Explanation:
From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.
- It could have emitted any form of radioactive particles which can be alpha or beta.
- We do not know if it has a long or short half life because the value is not given.
- But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
- A decay series involves a radioactive decay in multiple steps.
Cell division is the process by which a parent cell divides into two or more daughter cells. Cell division usually occurs as part of a larger cell cycle
