(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
They or on the southern hysteric watch it on youtube (latitude and longitude song 1 direction remix by the history teachers)
Answer:
Equilibrium concentrations of the gases are
Explanation:
We are given that for the equilibrium
Temperature, 
Initial concentration of
We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of
At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values
By solving we get
Now, equilibrium concentration of gases
Answer:
0.025 L
Explanation:
The production of oxygen in the electrolysis of water is;
4OH^-(aq) -----> 2H2O(l) + O2(g) + 4e
Since 1 F = 96500C
molar volume of a gas = 22.4 L
From the reaction equation;
4 * 96500 C yields 22.4 L of oxygen
(3 * 60 * 60 * 0.0400) C yields (3 * 60 * 60 * 0.0400) * 22.4/4 * 96500
= 9676.8/386000
= 0.025 L
