Answer:
1.196 M NaOH
Explanation:
Molarity = moles/Volume (L)
moles NaOH = mass NaOH/MM NaOH = 12/40.01 = 0.299 moles NaOH
Volume solution = 250 mL = 0.250L
M = 0.299/0.250=1.196 M NaOH
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Answer:
58.44 g of NaCl are needed.
Explanation:
Given data:
Mass of NaCl needed = ?
Volume of solution = 200 mL (200/1000 =0.2 L)
Molarity of solution = 5 M
Solution:
We will solve this problem through molarity formula.
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will put the values.
5 M = moles of solute / 0.2 L
Moles of solute = 5 mol/L × 0.2 L
Moles of solute = 1 mol
Mass of sodium chloride:
Mass = number of moles × molar mass
Mass = 1 mol × 58.44 g/mol
Mass = 58.44 g
Thus, 58.44 g of NaCl needed.
Answer:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Regards!