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Alex_Xolod [135]
3 years ago
13

β-Myrcene, a key intermediate in the production of several fragrances, is 7-methyl-3-methylene-1,6-octadiene.Draw the structure

of this triene.
Chemistry
1 answer:
sdas [7]3 years ago
3 0

Answer:

this is the structure of beta myrcene but im notsure which structure you were looking for

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Question 11. Identify the reducing agent <br> Sn+2 + AG 0 —&gt; Sn0 + Ag+
Temka [501]

Answer:

Ag 0 is the reducing agent.

Explanation:

Reducing -> gaining electrons

Oxidizing -> losing electrons

Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.

4 0
2 years ago
Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.
vagabundo [1.1K]

Answer:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Explanation:

Hello.

In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:

3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)

Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Best regards.

8 0
3 years ago
Determine the number of grams of carbon-14 remaining in this sample at 17,100 years.
earnstyle [38]

Answer: I need help

Explanation:

8 0
3 years ago
Write a skeleton equation for the reaction in which aqueous sodium chloride
IRISSAK [1]

Answer:

Explanation:according to question:

. Nacl (aq) + AgNO3 (aq) --> AgCl

(s) + NaNO3 (aq).balanced.

5 0
3 years ago
How many grams of Al(OH)3 are produced from 3.00 g of AlCl3 with excess of NaOH?
Blizzard [7]

Answer:

1.772 gram is the approximate answer

Explanation:

molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole

the reaction is

AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl

from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3

implies 132g AlCl3 gives 78g Al(OH)3

Implies 3g AlCl3 gives

3*122/78 = 1.772 grams

3 0
3 years ago
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