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>
Answer:
B
Explanation:
to conduct scientific exploration of the moon
V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?
By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K
The molarity of KOH is 0.1055 M
<u><em> calculation</em></u>
Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH
H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O
step 2: find the moles of H₂C₂O₄.2H₂O
moles = mass÷ molar mass
from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol
= 0.2000 g ÷ 126 g/mol =0.00159 moles
step 3: use the mole ratio to calculate the moles of KOH
H₂C₂O₄.2H₂O : KOH is 1:2
therefore the moles of KOH =0.00159 x 2 = 0.00318 moles
step 4: find molarity of KOH
molarity = moles/volume in liters
volume in liters = 30.12/1000=0.03012 L
molarity is therefore = 0.00318/0.03012 =0.1055 M
