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)
True
Pressure is the force per unit area on an object. The force in the case of air pressure is the weight of the air molecules; therefore, the air pressure is essentially due to gravity pulling down air molecules.
I believe the answer is cheamical.
Hope this helps
Answer:
Final volume=V₂ = 216.3 mL
Explanation:
Given data:
Initial volume = 120.0 mL
Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)
Final volume = ?
Final temperature = 197.0 °C (197+273 = 470 K)
Solution:
We will apply Charles Law to solve the problem.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 120 mL × 470 K /260.7K
V₂ = 56400 mL.K /260.7K
V₂ = 216.3 mL
Explanation:
To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.
Based on this premise, the number of moles of species on the reactant and product side must be the same;
Li + Br₂ → LiBr
Put a,b and c as the coefficient of each species
aLi + bBr₂ → cLiBr
balancing Li;
a = c
balancing Br;
2b = c
let a = 1;
c = 1
b = 
or a = 2, b = 1 , c = 2
2Li + Br₂ → 2LiBr
P + Cl₂ → PCl₃
Using the same method;
aP + bCl₂ → cPCl₃
balancing P;
a = c
balancing Cl;
2b = 3c
let a = 1;
c = 1
b = 
or
a = 2, b = 3, c = 2
2P + 3Cl₂ → 2PCl₃
iii,
H₂ + SO₂ → H₂S + H₂O
use coefficients a,b,c and d;
aH₂ + bSO₂ → cH₂S + dH₂O
balancing H;
2a = 2c + 2d
balancing S;
b = c
balancing O
2b = d
let b = 1,
c = 1
d = 2
a = 3
3H₂ + SO₂ → H₂S + 2H₂O
