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nikklg [1K]
3 years ago
9

Which factors must be equal when a reversible chemical process reaches equilibrium?

Chemistry
1 answer:
Anon25 [30]3 years ago
3 0
The concentration of the reactants and products remain constant. Because the rates of the forward and reverse reaction are equal there is no net change to the amount of reactants or products produced.May 19, 2011
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A student plans a two-step synthesis of 1-ethyl-3-nitrobenzene from benzene. The first step is nitration of benzene to give nitr
arlik [135]

Answer:

Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

Explanation:

The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.

-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.

This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.

7 0
3 years ago
51.7ml at 27 Celsius and 90kpa to stp
never [62]
STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K*  90^10^3Pa/ 300.15K *  1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml
6 0
3 years ago
What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
BARSIC [14]

Answer:

Mean

Explanation:

The mean of a series of measurements is calculated when a<em>ll the measurements are added up and then divided by the number of measurements taken</em>, as follows:

  • Sum of Measurements = 20.73 + 20.76 + 20.68 + 20.75 = 82.92

As<u> there are 4 measurements</u>, the mean is:

  • Mean = 82.92 / 4 = 20.73
3 0
2 years ago
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