I got that pH=3.65 using the fact that Ka=[H⁺][A⁻]/[HA] at equilibrium. In the ice table, I stands for initial, C stands for change, and E stands for equilibrium.
I hope this helps. Let me know if anything is unclear.
What happens is it makes water
The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.
Answer:
Chlorine gas.
Explanation:
Hello!
In this case, the undergoing chemical reaction is:
Thus, given the moles of reacting both sodium and chlorine, we compute the moles of sodium chloride yielded by each reactant by considering the 2:2 and 1:2 mole ratios:
Thus, since chlorine yields less moles of sodium chloride, we infer it is the limiting reactant.
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Answer:
Explanation:1. NaNH2 (1-Butene)
CH3CH2CH2CH2Cl --------------> CH3CH2CH=CH2 + HCl (elimination reaction)
2. Br2, CCl4
CH3CH2CH=CH2 ---------------> CH3CH2CH(Br)CH2Br (Simple addition Reaction)
3. NaNH2 (1-Butyne)
CH3CH2CH(Br)CH2Br ----------------> CH3CH2C≡CH + 2HBr
Sodamide (NaNH2) is a very strong base and generally results in Terminal Alkynes when treated with Vicinal Dihalides.
Alcoholic KOH on the other hand results in the formation of Alkynes with triple bonds in the middle of the molecule.
