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Zepler [3.9K]
3 years ago
7

draw the resonance structure of the three possible carbocation intermediate to show how methoxyl (-OCH3) diects bromination towa

rd ortho and para position​​
Chemistry
1 answer:
yanalaym [24]3 years ago
5 0

Answer:

See explanation and image attached

Explanation:

We know that some substituted benzene reacts faster than benzene towards electrophillic substitution. This is often due to the activation of the benzene ring towards electrophillic substitution by resonance.

-OCH3 directs an incoming electrophile (such as during bromination) to the ortho or para position. This is made possible by resonance (mesomeric) effect as shown in the image attached to this answer.

Image credit: pinterest

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How many grams of H 2 O can be produced with 6.3 moles of O 2 ?
il63 [147K]

Answer:

3.15×18

56.7

................

3 0
3 years ago
determine the final temperature of a gold nugget parentheses Mass equals 376 G parentheses that starts at 398k and loses 4.85 KJ
maw [93]

Answer:

297.23 K

Explanation:

Given that:

the mass of the gold nugget (m) = 376 g

the initial temperature T_1 = 398 K

amount of heat lost Q = -4.85 kJ = -4.85 × 10³

specific heat capacity (c) = 0.128 J/g° C

Using the formula for calculating the Heat energy

Q = mc\Delta T \\  \\  Q = m\times c\times (T_2-T_1)

-4.85 \times 10^3 = 376 \times  0.128 \times (T_2 - 398) \\ \\ -4.85\times 10^3 =  48.128(T_2 - 398) \\ \\ \dfrac{-4.85\times 10^3}{48.128}= (T_2 - 398) \\ \\-100.77 = T_2 -398\\\\T_2 = -398 +100.77\\ \\ \mathbf{ T_2 = 297.23K}

7 0
3 years ago
When a kettle is placed on the stove and water begins to boil, the hotter water at the bottom begins to rise and the cooler wate
melisa1 [442]
I think the correct answer from the choices listed above is option A. The fact that we know about the relationship between the hotter water and the cooler water would be that the <span> hotter water is less dense. Hope this answers the question. Have a nice day.</span>
3 0
3 years ago
Read 2 more answers
Consider a solution that is 2.5×10−2 M in Fe2+ and 1.1×10−2 M in Mg2+. (Ksp for FeCO3 is 3.07×10−11 and Ksp for MgCO3 is 6.82×10
Bumek [7]

Answer:

This question is incomplete, here's the complete question:

Consider a solution that is 2.1×10−2 M in Fe2+ and 1.6×10−2 M in Mg2+.

Part A

If potassium carbonate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first? ANSWER: Fe 2+

Part B

What minimum concentration of K2CO3 is required to cause the precipitation of the cation that precipitates first? ANSWER: [K2CO3] = 1.5×10−9 M

Part C

What is the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate? (ANSWER IS NOT .021 or 2.0E-6)

Explanation:

Part A

Fe2+ will precipitate first as solubility product of FeCO3 is lesser than solubility product of MgCO3.

Part B

FeCO3\= Fe2+ + CO32-

Ksp = [Fe2+][CO32-] = 3.07 x 10-11

3.07 x 10-11 = (2.1 x 10-2)[CO32-]

[CO32-] = 1.5 x 10-9 M to precipitate the Fe2+ ions

Part C

MgCO3\= Mg2+ + CO32-

Ksp = [Mg2+][CO32-] = 3.07 x 10-11

6.82 x 10-6 = (1.6 x 10-2)[CO32-]

[CO32-] = 4.3 x 10-4 M to precipitate the Mg2+ ions

Ksp = [Fe2+][CO32-] = 3.07 x 10-11

3.07 x 10-11 = [Fe2+](4.3 x 10-4)

[Fe2+] = 7.1 x 10-8 M when the Mg2+ ions precipitates

5 0
3 years ago
An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin
dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

2NH_3(aq)+CO_2 →CH_4N_2O(aq)+H_2O (l)

Molar mass of urea CH_4N_2O= 60g/mole

Moles of NH_3 = \frac{62.8kg/mole}{17g/mole} (since Moles=\frac{mass  of  substance}{mass of one mole})

                     = 4011.76 moles

Moles of CO_2 = \frac{105kg}{44g/mole}

                = \frac{105000g}{44g/mole}

                = 2386.36 moles

Theoritically , moles of NH_3 required = double the moles of CO_2

    but , 4011.76 , the limiting reagent is NH_3

Theoritical moles of urea obtained = \frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3

                                                      = 2005.88mole CH_4N_2O

Mass of 2005.88 mole of CH_4N_2O =2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}

                                                     = 120352.8g

                                                     120352.8g\times \frac{1kg}{1000g}

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \frac{87.5kg}{120.35kg}\times100

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0
3 years ago
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