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alexandr402 [8]
2 years ago
8

Given that the density of Hg(l) at 0°C is about 14 g mL-­1, which of the following is closest to the volume of one mole of Hg(l)

at this temperature?
Chemistry
1 answer:
Mila [183]2 years ago
8 0

Answer: 14mL

Explanation:

From the question, we've been told that the that the density of Hg(l) at 0°C is about 14 g mL-­1.

14g = 1mL

We should note that atomic weight of mercury = 200.59.

Therefore, 1 mol Hg = 200.59 g

Therefore, the answer that is closest to the volume of one mole of Hg(l) at this temperature will be:

= 200.59/14

= 14.3 mL

= 14mL approximately

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larisa86 [58]
Rust on a rock that contains iron
Mineral absorbs water and changes into a new substance
5 0
2 years ago
ammonium bromide reacts with copper(i) sulfate in aqueous solution to form a solid precipitate. what is the net ionic equation f
sukhopar [10]

Answer:

Br^-(aq)+Cu^+(aq)\rightarrow CuBr(s)

Explanation:

Hello,

In this case, the molecular chemical reaction is written as:

2NH_4Br(aq)+Cu_2S(aq)\rightarrow 2CuBr(s)+(NH_4)_2SO_4(aq)

In such a way, considering the net ionic notation and the solid yielded copper (I) bromide, we write:

2NH_4^+(aq)+Br^-(aq)+Cu^+(aq)+SO_4^{2-}(aq)\rightarrow CuBr(s)+NH_4^+(aq)+SO_4^{2-}(aq)

Whereas ammonium and sulfate ions are spectator ions, therefore the net ionic equation is:

Br^-(aq)+Cu^+(aq)\rightarrow CuBr(s)

Regards.

5 0
2 years ago
An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t
Elenna [48]

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol

Thus, the concentration of ethylamine in solution is:

[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M

Now, we can also infer that some salt is formed, and has the following concentration:

[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0

Finally, the pH turns out to be:

pH=14-pOH=14-3\\\\pH=11

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

4 0
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GenaCL600 [577]

Answer:

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Explanation:

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<span> full moon is the best answer, but perhaps 90% lit, not entirely. The features we see on a full moon are shadows = not lit areas.</span>
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