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Lisa [10]
4 years ago
10

substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. substituents can act

ivate or deactivate the ring to substitution, donate or withdraw electrons inductively, donate or withdraw electrons through resonance, and direct substitution either to the ortho/para or to the meta positions. from the following lists, select the substituents that have the indicated property. the substituents are written as -xy, where x is the atom directly bound to the aromatic ring.Activation of the ring towards substitutionCH3 NH2 NO2 None of above
Chemistry
1 answer:
Tamiku [17]4 years ago
3 0

Answer:

Ortho para directed substitution------------------     CH3

Explanation:

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What is the volume of 3.40 mol of gas at 33.3C and 22.2atm of pressure
Luda [366]
The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
<span>The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.</span>
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3 years ago
22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​
IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

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