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nlexa [21]
3 years ago
14

Draw the product(s) obtained when benzoquinone is treated with excess butadiene. using wedges and dashes, indicate the stereoche

mistry.

Chemistry
1 answer:
Alex17521 [72]3 years ago
6 0

Answer:

The structure can be found on the attached documents

Explanation:

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Consider the following reversible reaction.
andriy [413]

Answer:

No one is correct. The correct expression is:

Keq = [H₂]²  . [O₂]² / [H₂O]²

Explanation:

To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.

The balance reaction is:

<u>2</u> H₂O (g)  ⇄  <u>2</u> H₂ (g)  +  O₂ (g)

Keq = [H₂]²  . [O₂]  / [H₂O]²

In opposite side: <u>2</u> H₂ (g)  +  O₂ (g)   ⇄  <u>2</u> H₂O (g)

Keq =  [H₂O]² / [H₂]²  . [O₂]  

6 0
4 years ago
In which region are most particles moving the fastest?
tester [92]
Container which is heated
3 0
3 years ago
Read 2 more answers
as 100 milliliters of 0.10 molar KOH is added to 100 milliliters of 0.10 molar HCL at 298 K, the pH of the res?A. decrease to 3B
Mila [183]

Answer:

C. increase to 7.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

KOH+HCl\rightarrow KCl+H_2O

Thus, the molar relationship is 1 to 1, therefore, the moles are:

n_{HCl}=0.1mol/L*0.1L=0.01molHCl\\n_{KOH}=0.1mol/L*0.1L=0.01molKOH\\

Thus, since the entire hydrogen ions are neutralized, the pH C. increase to 7.

Best regards.

8 0
3 years ago
In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc
natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




3 0
3 years ago
Answer please????? Thank you
snow_tiger [21]
Can’t have anything in Detroit
7 0
3 years ago
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