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pav-90 [236]
3 years ago
5

What specific electrophile does benzene attack when it undergoes nitration??

Chemistry
1 answer:
Deffense [45]3 years ago
7 0
<span><span>The reaction is as follows:
        C6H6   </span>+   HNO3   +      H2SO4     ------------> </span>C6H5NO2<span>   +   H</span>2<span>O
(BENZENE) (NITRIC ACID)(CATALYST)
</span>NO2(+) is the electrophile that acctacks on the benzene ring in nitration process.
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separating a solid from a liquid by evaporating is called _____ a) filtration b) condensation c) solution d) distillation
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Calculate the energy required to heat 322.0g of ethanol from −2.2°C to 19.6°C . Assume the specific heat capacity of ethanol und
just olya [345]

Answer:

There is 17.1 kJ energy required

Explanation:

Step 1: Data given

Mass of ethanol = 322.0 grams

Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K

Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K

Specific heat capacity = 2.44 J/g*K

Step 2: Calculate energy

Q = m*c*ΔT

⇒ m = the mass of ethanol= 322 grams

⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K

⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K

Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ

There is 17.1 kJ energy required

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3 years ago
Weight is measured using an instrument called a
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I believe the answer is a balance or mechanical scale
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Using the ideal gas law, PV=nRT, where R=0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic
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Answer:

V = 2.32 Liters

Explanation:

PV = nRT => V = nRT/P

n = 25.8g/122g/mole = 0.21 mole

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T = 25.44°C + 273 = 298.44K

P = 2.22 atm (given in problem)

V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm

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A negatively charged particle is attracted to A. All particles that are located close by. B. Only particles that are large. C. N
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Answer:

D. Positively charged particles

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Negatively charged particles are attracted to positively charged particles and repelled against negatively charged particles

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