Answer:
Equilibrium constant Kc = Qc = quotient of reactant(s) and product(s)
Kc = [C]x[D]y..../[A]m[B]n..... = 0.328dm3/mol, where [C]x[D]y is the product and [A]m[B]n is the reactant(Both in gaseous states)
Explanation:
When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.
That a reaction quotient always assumes the same value at equilibrium can be expressed as:
Qc (at equilibrium) = Kc =[C]x[D]y…/[A]m[B]n…
This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.
Answer: they contain solutes and solvents, their particles must be evenly distributed, they may contain solid liquid and gas simultaneously, and they are homogeneous matter.
Explanation:
We can use the combined gas law equation to solve for the initial volume
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
V = 8.50 L
the initial volume is 8.50 L
Answer:
Where are the following?
Explanation:
The main properties of water are its polarity, cohesion, adhesion, surface tension, high specific heat, and evaporative cooling.
Answer:
B) -1551 kJ
Explanation:
There are two heat flows in this question.
Heat released by engine + heat absorbed by water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m = 2.51 kg
C = 3.41 J°C⁻¹g⁻¹
T_i = 23.8 °C
T_f =205 °C
Calculations:
(a) ΔT
ΔT = 205 °C - 23.8 °C = 181.2 °C
(b) q₂
q₂ = mCΔT = 2510 g × 3.41 J·°C⁻¹g⁻¹ × 181.2 °C = 1.551× 10⁶ J = 1551 kJ
(c) q₁
q₁ + 1551 kJ = 0
q₁ = -1551 kJ
