Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
Answer:
A. 4-ethyl-hex-3,5-dien-2-ol.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol.
Explanation:
Hello there!
In this case, according to the given problems, it is possible to apply the IUPAC rules to obtain the following names:
A. 4-ethyl-hex-3,5-dien-2-ol because we have an ethyl radical at the fourth carbon and the beginning of the parent chain is on the Me (CH3) because it is closest to first OH.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol: because we start at the alcohol and have a chlorine atom on the second carbon, a methyl radical on the third carbon, a <em>tert</em>-butyl on the fifth carbon and the parent chain is benzene which is phenol as an alcohol.
Regards!
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
Answer:
idk
Explanation:
idk cool pee bee mee nee hee gee fee kee
