First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Wym kingdom? I dont get it
If the enthalpy change is negative, then the reaction is exothermic because heat energy was lost to the surroundings.
If the enthalpy change was positive, then the reaction was endothermic because heat energy was gained from the surroundings!
Hope this helps!! x
Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
Iron(II) oxide also refers to a family of related non-stoichiometric compounds, which are typically iron deficient with compositions ranging from Fe0.84O to Fe0.95O.
