The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anodeto the cathode.
Answer:
0.414 mole (3 sig. figs.)
Explanation:
Given grams, moles = mass/formula weight
moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)
≅ 0.414 mole (3 sig. figs.)
Answer:
The biggest risk with recharging alkaline batteries is leakage. As you probably know, alkaline batteries leak even under normal circumstances. Internal off gassing, made worse by heat, creates pressure that can breach battery seals. Therefore, the risk of leakage is an even bigger risk when recharging.
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
= 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial 0.00525 0 0
change - 0.002625 +0.002625 +0.002625
equilibrium 0.002625 0.002625 0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
= -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
= 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
Answer:
C)
Explanation:
(A) Cover the chamber. => <u>It helps to keep the solvent in his gas state</u>
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(B) Gently swirl the solvent in the chamber prior to placing the TLC plate inside => <u>The mechanical energy can promote the conversion from the liquid state to the gas state and help to the saturation process.</u>
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(C) Add ninhydrin (a visualization technique) to the developing solvent =><u>There is no change in the saturation process</u>
(D) Place a paper wick (like a piece of filter paper) inside the chamber=><u>Increases the area for evaporation. The solvent can go up across the paper by capillarity a then can be evaporated increasing the saturation in the chamber.</u>
