You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
Answer:
The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L
Explanation:
Molarity (M) means: moles of solute which are contained in 1 L of solution.
In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.
0,2 moles ........... 1 L
5x10*-2 moles ........... x
x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L
(we can also say 250 mL)
Answer:
904.014 j/kgk
Explanation:
Mass of metal = 45g
Temperature of metal = 85.6°
Mass of water = 150
Temperature of water = 24.6
Final temperature of system = 28.3
Heat lost by metal = Heat gained by water
m1 * c1 * dt = m2 * c2 * dt
Q = quantity of heat
Q = m*c*dt
dt = change in temperature
dt of water = 28.3 - 24.6 = 3.7
dt of metal = 85.6 - 28.3 = 57.3
Specific heat capacity of water, c = 4200
(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7
2.5785c1 = 2331
c1 = 2331 / 2.5785
= 904.01396
= 904.014 j/kgk
Answer:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
Explanation:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
