Answer:
: cathode rays
a beam of electrons emitted from the cathode of a high-vacuum tube.
During the 1880s and '90s scientists searched cathode rays for the carrier of the electrical properties in matter. Their work culminated in the discovery by English physicist J.J. Thomson of the electron in 1897.
Answer:
2p
Explanation:
it has 3 dumbell shapes, hence p
you can't determine the principal quantum number by looking at the shape, however bigger or spread orbital means higher value of n
Answer:
See Explanation
Explanation:
a. pH of 1M HOAc(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 1.0M x x
Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵
=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M
=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37
b. pH of 0.10M CH₃NH₃OH(aq)
CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴
C(eq) 0.10M x x
=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M
=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M
=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18
=> pH = 14 – pOH = 14 – 2.18 = 11.82
c. pH of 0.30M HOAc/0.10M OAcˉ(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 0.30M x 0.10M
=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]
= 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M
=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26