<span><span>The reaction is as follows:
C6H6 </span>+ HNO3 + H2SO4 ------------> </span>C6H5NO2<span> + H</span>2<span>O
(BENZENE) (NITRIC ACID)(CATALYST)
</span>NO2(+) is the electrophile that acctacks on the benzene ring in nitration process.
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
We mostly use Kilograms and grams to measure mass =)
Explanation:
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
xoxo hope this helps
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