Answer : The complete net ionic reaction between
![MnO _{4} ^{-}](https://tex.z-dn.net/?f=MnO%20_%7B4%7D%20%20%5E%7B-%7D%20)
and
![Fe ^{2+}](https://tex.z-dn.net/?f=Fe%20%5E%7B2%2B%7D)
in an acid solution, can be written as :-
Explanation:
(a) a functional group is a special group of atoms or bonds in a compound that is responsible for the chemical reactions, behavior, and characteristics of that compound. (b) Functional group is present in both the compound is alcohol (-OH).
Functional groups are specific groupings of atoms within molecules that have their own characteristic properties, regardless of the other atoms present in a molecule. Common examples are alcohols, amines, carboxylic acids, ketones, and ethers.
Key Points
Functional groups are often used to “functionalize” a compound, affording it different physical and chemical properties than it would have in its original form.
Functional groups will undergo the same type of reactions regardless of the compound of which they are a part; however, the presence of certain functional groups within close proximity can limit reactivity.
Functional groups can be used to distinguish similar compounds from each other.
Key Terms
functional group: A specific grouping of elements that is characteristic of a class of compounds, and determines some properties and reactions of that class.
functionalization: Addition of specific functional groups to afford the compound new, desirable properties.
The limiting reagent is Copper (Cu)
First, balance the reaction before determining the limiting reactant.
Cu + 2Ag
Cu(NO3)2 + 2Ag from NO3
Take the beginning material amount and calculate the number of moles of one of the products to identify the limiting reactant. The limiting reactant is the one that produces a product with the fewest moles.
5.50×1.00/2.00=2.75
The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.
Copper is the limiting reactant since there are only 2.50 mol of it and only 2.50 mol of AgNO3 will result from the reaction.
Learn more about the limiting reagent brainly.com/question/14795543
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Calories is a unit of energy, so we need to calculate the energy or heat absorbed or released by the system. We use the equation:
U = mCpΔT
where:
m = mass of material
Cp = specific heat capacity of material
ΔT = (Tf - Ti) = temperature change
First, you must look for the specific heat capacity of copper. You can use any chemistry or engineering book or website. According to Engineering Toolbox, the specific heat capacity of copper is 0.09 cal/(g-°C).
U = 200(0.09)(33-28) = 90 calories
Since the value is positive, energy is absorbed.
Answer:- The Ka for the acid is
.
Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
Now, we make the ice table for this equation as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
I 0.25 0 0
C -X +X +X
E (0.25 - X) X X
where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.
X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.
![Ka = [H^+][A^-]\frac{1}{HA}](https://tex.z-dn.net/?f=Ka%20%3D%20%5BH%5E%2B%5D%5BA%5E-%5D%5Cfrac%7B1%7D%7BHA%7D)
Where, Ka is the acid ionization constant. Let's plug in the values.
![Ka = \frac{X^2}{0.25-X}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7BX%5E2%7D%7B0.25-X%7D)
Let's calculate the value of X first using the equation:
[/tex]
on taking antilog ob above equation we get:
![[H^+]=10^-^p^H](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5Ep%5EH)
![[H^+]=10^-^2^.^7^1](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5E2%5E.%5E7%5E1)
= 0.00195
So, X = 0.001195
Let's plug in this value of X in the equation:-
![Ka=\frac{(0.00195)^2}{0.25-0.00195}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%280.00195%29%5E2%7D%7B0.25-0.00195%7D)
![Ka=1.53*10^-^5](https://tex.z-dn.net/?f=Ka%3D1.53%2A10%5E-%5E5)
So, the value of Ka for butyric acid is
.