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JulijaS [17]
3 years ago
15

Bromine (Br,) is observed as a red-brown color During the bromination of benzene, an electrophilic aromatic substitution, why do

es the red-brown color of Br2 disappear as the reaction progresses?
a. The bromine (Br2) dissolves in the solution and becomes colorless
b. The bromine (Br2) has reacted with the benzene, leaving fewer bromine molecules in solution.
c. The bromine (Br,) reacts with acid to produce HBr, which is colorless .
d. The bromine (Br,) leaves the mixture as bromine gas .
Chemistry
1 answer:
sergey [27]3 years ago
4 0

Answer:

b. The bromine (Br2) has reacted with the benzene, leaving fewer bromine molecules in solution.

Explanation:

Colour is caused by absorption of light of the frequency of “complementary colours”. Exactly which frequencies molecules absorb depends on the arrangement of electrons. Obviously, to break the Br-Br bond and the C=C bond in the reactants and replace them with two C-H bonds is a significant reordering of electrons and so the new molecule is no longer able to absorb the same frequencies as previously. Hence, a colour change, in this case from orange to colourless.

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Phosphofructokinase catalyzes the phosphorylation of fructose 6‑phosphate to fructose 1,6‑bisphosphate in glycolysis. Fructose 1
umka2103 [35]

Answer:

d. increases PFK activity, decreases FBPase activity

Explanation:

Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.

Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.

Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate

3 0
3 years ago
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m
IrinaK [193]

Answer:

49.45~%

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

C_6H_1_2O~->~C_6H_1_0~+~H_2O

So, if we start with <u>10 mol of cyclohexanol</u> (C_6H_1_2O) we will obtain 10 mol of cyclohexanol (C_6H_1_0). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

(6*12)+(10*1)=82~g/mol

With this value we can calculate the grams:

10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0

Now, we have as a product 500 mL of C_6H_1_0. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

500~mL\frac{0.811~g}{1~mL}=405.5~g

Finally, with these values we can calculate the <u>yield</u>:

%~=~\frac{405.5}{820}x100~=~49.45%%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

6 0
3 years ago
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Travka [436]

A B and C are all chemical changes. paper tearing is not.

6 0
3 years ago
How many moles of Ca(OH)2 are in 3.5kg of Ca(OH)2? Answer in units of mole
Kobotan [32]
First, you need to convert kg to g. 
So, 1 kg =1000g.
3.5 x 1000 = 3500g Ca(OH)2

We need to know the molar mass of Ca(OH)2. 
Ca= 40.08 g
O=2(15.999)
H=2(1.0079)

Add them all together and you get 74.0938 g.

Put it in the formula from mass to moles. 

# of moles = grams Ca(OH)2 x 1 mol Ca(OH)2
                                                  --------------------
                                                  molar mass Ca(OH)2

3500 g Ca(OH)2 x 1 mol Ca(OH)2
                              ---------------------
                             74.0938 g Ca(OH)2

So divide 1/74.0938 and multiply by 3500.

You will get about 47.24 moles Ca(OH)2.

Hope this helps! :)
7 0
3 years ago
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