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3 years ago

For the following reaction in aqueous solution, identify all those species that will be spectator ions. Select all that apply.

Chemistry

1 answer:

uranmaximum [27]3 years ago

7 0

The reaction is incorrect. The correct reaction is: Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃. The options are:

A. Hg₂SO₄

B. Na₂SO₄

C. Na⁺

D. NO₃⁻

E. SO₄⁻²

F. Hg₂(NO₃)₂

G. NaNO₃

H. Hg⁺²

Answer:

C, D, E, H

Explanation:

The ions are formed after dissociation of the compound in the solution. When they're negatively charged, they're called anions, when they're negatively charged, they're called cations. If the ion is presented on both sides of the reaction, it is called a spectator ion.

Thus, the reaction given:

Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃

So, let's do the dissociation.

Na₂SO₄ is formed by the ions Na⁺ and SO₄⁻²;

Hg(NO₃)₂ is formed by the ions Hg⁺² and NO₃⁻;

HgSO₄ is formed by the ions Hg⁺² and SO₄⁻²;

NaNO₃ is formed by the ions Na⁺ and NO₃⁻.

Thus, the spectator ions are Na⁺, SO₄⁻², Hg⁺², and NO₃⁻.

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3 years ago

Metric conversions. Please help ASAP.

lesya692 [45]

Answer:

14. 13.2cg = 1.32dg

15. 3.8m = 0.0038km

16. 24.8L = 24800mL

17. 0.87kL = 870L

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19. 0.001hm = 10cm

Explanation:

14. 13.2/10 = 1.32

15. 38/1000 = 0.0038

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17. 0.87(1000) = 870

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19. 0.001hm(10000) = 10

An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.

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3 years ago

The reaction of pyrrole with bromine forms predominantly __________. View Available Hint(s) The reaction of pyrrole with bromine

xenn [34]

Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the resonance structures. (Figure 1), on these structures, we will obtain a negative charge on carbon 2 in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An electrophile is generated by the $Br_2$ and $FeBr_3$ . This electrophile can be attacked by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).