Answer:
Following are the solution to this question:
Explanation:
Please find the complete question in the attachment.
Start of Laboratory
Dissolve 2-naphthol in the round bottom flask with ethanol.
Add pellets of sodium hydroxide and hot chips. Attach a condenser.
Heat for 20 minutes under reflux, until the put a burden dissolves.
After an additional hour, add 1-Bromobutane and reflux.
Pour the contents into a beaker with ice from a round bottom flask.
On a Bachner funnel, absorb the supernatant by vacuum filtration.
Utilizing cold water to rinse the material and dry that on the filter.
Ending of the Lab
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
Answer:
20.3-17.5=2.8ml 1ml=1cm3 vol= 2.8cm3
Explanation:
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)
