5 types of air pollution would include: Carbon Monoxide, Lead, Nitrogen oxides, sulfur oxides. I only know those 4 actually… hope it helped
Balanced half-reactions:
Anode: Fe2+ -> Fe
Cathode: Mg2+ -> Mg
The cell is voltaic because the cathode has a more negative reduction potential, causing the cell potential to be positive.
The electrons flow from the anode to the cathode.
Answer:
158.2598
Explanation:
158.2598 moles are in 78.8 g of potassium sulfite
CARRY ON LEARNING
Answer:
(a). Increase in the rate of Reaction.
(b). Increase in the formation of the alcohol.
(c). rate of Reaction will increase that is to say more of the alkyl azide will be produced.
(d).rate of Reaction will increase that is to say more of the alkyl azide will be produced.
(e). rate of reaction increases because of carbocation.
Explanation:
Important thing to note is that this reaction proceed with an SN1 mechanism.
Therefore; the rate of reaction depends solely on the concentration of alkyl azide.
(a). The concentration of the alkyl chloride is increased: rate of Reaction will increase that is to say more of the alkyl azide will be produced.
(b) The concentration of the azide ion is increased: the effect is that there will be an increase in the formation of more of the alcohol.
c) The alkyl bromide is used instead of the alkyl chloride: the effect is that there will be rate of Reaction will increase that is to say more of the alkyl azide will be produced.
(d) The reaction is carried out in 1:1 acetone water instead of pure water: the effect is that there will be the effect is that there will be rate of Reaction will increase that is to say more of the alkyl azide will be produced.
(e) The alkyl chloride shown below is used instead of cumyl chloride: rate of reaction increases because of carbocation.
Answer:
30.0 L.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution = (MV)after dilution</em>
M before dilution = 5.0 M, V before dilution = 3.0 L.
M after dilution = 0.5 M, V after dilution = ??? L.
∵ (MV)before dilution = (MV)after dilution
∴ (5.0 M)(3.0 L) = (0.5 M)(V after dilution)
<em>∴ V after dilution = (5.0 M)(3.0 L)/(0.5 M) = 30.0 L.</em>
