Help please just #17

Chemistry

2 answers:

VARVARA [1.3K]2 years ago

5 0

The density of the unknown liquid is 0.981 g/mL

Your density triangle tells you that

D = m/V or density = mass/volume

So, D = (15.2 g)/(15.5 mL)

You divide numbers by numbers and units by units.

D = 15.2/15.5 g/mL =0.980 645 161 g/mL (by my calculator)

The answer can have no more than three significant figures, so you round off to three significant figures and get

D = 0.981 g/mL

defon2 years ago

4 0

So you see the triangle with the letters, look at what you're given. you have mass and volume solving for density. So you take the mass and divide it by the volume. 15.2/15.5 which your answer is 0.98 or 1.0 that's your density.

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A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

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Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.