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Evgen [1.6K]
3 years ago
7

Give the measurement following sig fig rules for tools. Do not forget the unit after your number measurement.​

Chemistry
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

Explanation:

You are not really helped by what is hold the liquid. Beakers and cylinders come on a lot of sizes. I hate to be crabby about things like that, but you really need to be aware that the question is slightly flawed (not your fault).

The beaker, you'd be like to get 1 sig digit. You have to be awfully careful about claiming more. So the and is 50 mL, but that mL is a guess and the 50 is not totally accurate, but what would you say the second digit is? 48 or 47? You don't really know. Maybe even 49.

The graduated cylinder is a little better. Read the bottom of the meniscus (the bottom of the 1/2 bubble). I think you can get 2 sig digs., so the answer is 36 mL. But everything also depends on what you have been told.

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The rate of the backward reaction increases

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3 years ago
Calculate the ph of the solution resulting from the addition of 85.0 ml of 0.35 m hcl to 30.0 ml of 0.40 m aniline (c6h5nh2). kb
stiv31 [10]

Answer:

pH = 0.81

Explanation:

HCl reacts with aniline, thus:

C₆H₅NH₂ + HCl → C₆H₅NH₃⁺ + Cl⁻

Moles of HCl are:

0.085L × (0.35mol HCl / L) = <em>0.02975mol HCl</em>

Moles of aniline are:

0.030L × (0.40mol HCl / L) = <em>0.012mol aniline</em>

Thus, after reaction, will remain:

0.02975mol - 0.012mol = <em>0.01775mol HCl</em>

Moles of HCl in solution are equal to moles of H⁺, thus, moles of H⁺ are: <em>0.01775mol H⁺</em>

As total volume is 85.0mL + 30.0mL = 115.0mL ≡ <em>0.115L</em>

0.01775mol / 0.115L = 0.1543M

pH of solution = -log[H⁺]

pH = -log 0.1543M

<em>pH = 0.81</em>

<em></em>

3 0
3 years ago
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