Answer:
Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture
Explanation:
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Answer:
C = 9460 Kj
Explanation:
Given data:
Mass of copper = 2kg
Latent heat of vaporization = 4730 Kj/Kg
Energy required to vaporize 2kg copper = ?
Solution:
Equation
Q= mLvap
by putting values,
Q= 2kg × 4730 Kj/Kg
Q = 9460 Kj
Cheetah who is running with speed of 80 miles per hour then it is his instantaneous speed becaue i shows as in per hour speed..... hope it helps