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Sergeu [11.5K]
3 years ago
8

4. Which of the following aromatic compounds undergoes nitration faster than benzene does? Explain your answer

Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

Ph-OH

Explanation:

Nitration of Aromatic compounds is an example for Aromatic Electrophilic substitution and its rate is enhanced by Electron donating group like OMe, OH, alkyl group and halides (least activating group) while reactivity is diminished by electron withdrawing like CN,NO2 and Ketone etc

Hence phenol is more reactive than benzene towards nitration.

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How many moles of NaOH are required to prepare 2.90 L of 1.8 M NaOH?
marusya05 [52]
V=2.90 L
c=1.8 mol/L

n(NaOH)=vc

n(NaOH)=2.90L*1.8mol/L=5.22 mol
3 0
3 years ago
Read 2 more answers
Convert the following to Celsius
Naya [18.7K]

9. (80-32)5/9

48×5/9

240/9

26.66°C

9. (90-32)5/9

58×5/9

32.22°C

10 (212-32)5/9

180×5/9

20×5

100°C

8 0
3 years ago
What is the term for a liquid composed of polar molecules?
IrinaVladis [17]
A dissolving liquid composed of polar molecules is a polar solvent.

The distinction of polar and non-polar liquids is important because the like dissolves like rule. This rule states that the solubility is greater when the polarity of the liquid is similar to the polarity of the solute.

So, to dissolve polar compounds (e.g. ionic compounds) you should use polar solvents (e.g. water).

Answer: polar solvent
7 0
3 years ago
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
C What mass of gas is present in<br>48 cm of oxygen. O2​
xenn [34]

Answer:

carbon dioxide

Explanation:

because we breath in oxygen and breath out Carbon

7 0
3 years ago
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