Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
Answer:
b. transfer of electron(s).
Explanation:
An oxidation-reduction also called a redox reaction is a
chemical reaction in which electrons are transferred of between two species of reactants. It is a chemical reaction where the oxidation number of an atom, ion, or molecule, increases or decreases by losing or gaining electrons
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
Explanation:
Given parameters:
Number of molecules = 4.21 x 10²³ molecules
Unknown parameters:
Number of moles
Solution:
A mole can be defined as the amount of a substance that contains the avogadro's number of particles i.e 6.02 x 10²³
To find the number of moles:
Number of moles = ![\frac{number of molecules }{[tex]6.02 x 10^{23}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20of%20molecules%20%7D%7B%5Btex%5D6.02%20x%2010%5E%7B23%7D)
}[/tex]
Number of moles = ![\frac{4.21 x 10[tex]^{23}](https://tex.z-dn.net/?f=%5Cfrac%7B4.21%20x%2010%5Btex%5D%5E%7B23%7D)
}{
}[/tex]
Number of moles of CaCl₂ = 0.699moles
Learn more;
mole calculation brainly.com/question/13064292
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