Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
Answer:
<h3>I don't know what is the answer of your question sorry never mind..</h3>
Explanation:
<h3>And please marks me as brainliest... </h3>
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the
substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
Density = mass / volume
The density of water = 0.997 g/mL
<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>
<span> = 1246.25 g</span>
Specific heat capacity of water = 4.186 J<span>/ g °C.</span>
Let's assume that there is no heat loss to the surrounding and the final temperature is T.
By applying the equation,
5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C
Hence, the final temperature of the water is 24.04 °C.
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