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Sonbull [250]
3 years ago
14

Assessment timer and count

Chemistry
2 answers:
ollegr [7]3 years ago
8 0

Answer:

The answer to your question would be "the insect touching the trigger hairs".

Explanation:

I'm not much of an ex-plainer but I know this is the right answer because I took the test and got a 100%. Please trust me on this. If wrong, please tell me. This is what I was taught at school. Thank you and good day.

love history [14]3 years ago
4 0

Answer:

The answer is: "the insect touching the trigger hairs"

Explanation:

I took the k-12 test and got a 100%!

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Ka for acetic acid, CH3COOH, is 1.80×10-5.Ka for acetylsalicylic acid (aspirin), HC9H7O4, is 3.00×10-4.Ka for benzoic acid, C6H5
klemol [59]

Answer:

The weakest conjugate base is C9H7O4⁻ which is the conjugate base of Aspirin.

Explanation:

The higher the Ka, the stronger the acid thus the weaker its conjugate base.

Ka of the acid are provided:

CH3COOH = 1.80*10-5

HC9H7O4 = 3.00*10-4

C6H5COOH = 6.30*10-5

From the Ka above, HC9H7O4 have the highest therefore its conjugate acid is the weakest.

When it donates the proton, it becomes the conjugate base.

HC9H7O4 + H2O ⇄ C9H7O4⁻ + H3O⁺

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