Answer:
The formula of Organic acid is as follow,
R-COOH
Explanation:
The class of organic acids is called Carboxylic Acids. In above general structure, R is alkyl group and can vary. While -COOH is the functional group.
Carboxylic Acids has the tendency to loose protons and their pKa value depends upon the alkyl group. For example the pKa value of Acetic acid (R = -CH₃) is 4.7. The driving force for this acidity is the stability of carboxylate (conjugate base) due resonance. i.e
RCOOH ⇄ RCOO⁻ + H⁺
Where;
RCOO⁻ = Carboxylate Ion (Conjugate base)
Answer : The concentration of 
at equilibrium is 0 M.
Solution : Given,
Concentration of = 0.0200 M
Concentration of 
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of 
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)
By solving the term, we get:
Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of at equilibrium is 0 M.
Explanation:
Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.
A. One of the oxides (Oxide 1) contains 63.2% of Mn.
Mass of the oxide = 100g
Mass of Mn = 63.2 g
Mass of O = 100 - 63.2
= 36.8 g
Ratio of Mn to O = 63.2/36.8
= 1.72
Another oxide (Oxide 2) contains 77.5% Mn.
Mass of oxide = 100 g
Mass of Mn = 77.5 g
Mass of O = 100 - 77.5
= 22.5 g
Ratio of Mn to O = 77.5/22.5
= 3.44
Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.
B.
Oxide 1
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Oxide 2
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Answer:
The material which is likely to slow the flow of electric charges the most is PVC plastic ‾ \text{\underline{PVC plastic}} PVC plastic.
Explanation:
brainlist please
Answer:
S(s) + O2(g) --> SO2(g)
Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).
Explanation:
The reaction is given as;
Sulfur + oxygen --> Sulphur dioxide
Sulphur = S
Oxygen = O2
Sulfur dioxide = SO2
So we have;
S(s) + O2(g) --> SO2(g)
The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).
