Realize that pH + pOH = 14
so, 9 + pOH = 14 -> pOH = 5
pOH = -log[OH-]
5 = -log[OH-]
plug it into a calculator and you get 1.0 x 10^-5
alternatively, use [OH-] = 10^-pOH to get the same answer
[OH-] = 1.0 x 10^-5
Answer:
410.196 J/[kg*°C].
Explanation:
1) the equation of the energy is: E=c*m*(t₂-t₁), where E - energy (523 J), c - unknown specific heat of copper, m - mass of this copper [kg], t₂ - the final temperature, t₁ - initial temerature;
2) the specific heat of copper is:
![c=\frac{E}{m*(t_2-t_1)}; \ => \ c=\frac{523}{0.085*(45-30)}=\frac{523}{1.275}=410.196[\frac{J}{kg*C}].](https://tex.z-dn.net/?f=c%3D%5Cfrac%7BE%7D%7Bm%2A%28t_2-t_1%29%7D%3B%20%5C%20%3D%3E%20%5C%20c%3D%5Cfrac%7B523%7D%7B0.085%2A%2845-30%29%7D%3D%5Cfrac%7B523%7D%7B1.275%7D%3D410.196%5B%5Cfrac%7BJ%7D%7Bkg%2AC%7D%5D.)
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
