To be able to write correctly the equilibrium expression of a reaction, we need to know the balanced reaction and the phases of the substances in the reaction. When substances are solid, pure liquid they are not included in the expression. We do as follows:
<span>4KO2(s) + 2H2O(g) = 4KOH(s) + 3O2(g)
K = [O2]^3 / [H2O]^2</span>
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
.
Note that water isn't part of this constant.
The value of
at 25 °C is
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
for pure water. - As a result,
at 25 °C.
Back to this question.
is given. 25 °C implies that
. As a result,
.
Answer:
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Explanation:
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Too much money and dangerous
Answer:
0.75 cal/g°c
Explanation:
for specific heat we have formula:
Amount of heat absorbed or released = mass x specific heat of a substance x change in temperature.
ΔQ=m x c x ΔT
where c= specific heat
m= mass of a substance
ΔT = total temperature
ΔQ = Amount of heat
so for specific heat,
c= ΔQ/mxΔT
c= 280/25x (25-10)
c= 280/375
c= 0.75 cal/g°c