4 NH3 + 6 NO → 5 N2 + 6 H2O How many moles of NH3 are necessary to produce 0.824 mol N2?

Sav [38]

Answer:

Sewage is a liquid containing wastes from households, industrial and agricultural activities discharged into water and it is dirty water.

7 0

2 years ago

Arrange the following in order of increasing bond strength of the carbon oxygen bond:

icang [17]

Answer: Option (d) is the correct answer.

Explanation:

It is known that length of a bond is inversely proportional to the bond strength. This also means that a single bond has long length due to which it is weak in nature.

And, a double bond is shorter in length and has more strength as compared to a single bond. Whereas a triple bond has the smallest length and it has high strength as compared to a double or single bond.

For example, carbon monoxide is CO where there is a triple bond between the carbon and oxygen atom.

Carbon dioxide is $CO_{2}$ where there exists a double bond between the carbon and oxygen atom.

A carbonate ion is $CO^{2-}_{3}$ when two oxygen atoms are attached through single bond with the carbon atom and another oxygen atom is attached through a double bond to the carbon atom.

Hence, we can conclude that order of increasing bond strength of the given carbon oxygen bond is as follows.

Carbonate ion < carbon dioxide < carbon monoxide

5 0

3 years ago

Question 3. A batch chemical reactor achieves a reduction in

kotykmax [81]

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

$r = k [A]^{x} [B]^{y}$

where:

[A] is the concentration of species A,
x is the order with respect to species A.
[B] is the concentration of species B,
y is the order with respect to species B

k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

$v(t) = -\frac{d[A]}{dt} = k [A]^{n}$

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law: rate = k

Concentration-time Equation: [A]=[A]o - kt

where

k: rate constant [M/s]

[A]: concentration in the time t [M]
[A]o: initial concentration [M]
t: elapsed reaction time [s]

For first-order kinetics, we have:

Rate Law: rate= k[A]

Concentration -Time Equation: ln[A]=ln[A]o - kt

where:

K: rate constant [1/s]
ln[A]: natural logarithm of the concentration in the time t [M]
ln[A]o: natural logarithm of the initial concentration [M]
t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use: [A]=[A]o - Kt

we replace the data: 5 = 100 - K (60)

we clear K: K = [100 - 5 ] (mg/L) /60 (s) = 1, 583 [mg/L.s]

First-order kinetics

we use: ln[A]=ln[A]o - Kt

we replace the data: ln(5) = ln(100) - K (60)

we clear K: K = [ln(100) - ln(5)] /60 (s) = 0,05 [1/s]

4 0

3 years ago

Calculate the molar mass of the acid.

JulijaS [17]

Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)

3 0

3 years ago

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$