Answer:
0.823 M was the molarity of the KOH solution.
Explanation:
(Neutralization reaction)
To calculate the concentration of base , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
0.823 M was the molarity of the KOH solution.
Answer:
The 2 would be placed in front of the reactant Na and in front of the product NaCl
2Na + Cl2 = 2NaCl
Explanation:
This is because the ratio of elements needs to be balanced on both sides.
On the reactants side, there are 2 Na molecules and 2 Cl molecules
On the products side, there are 2 Na molecules and 2 Cl molecules
So, now the equation is balanced
The answer would be Force
Answer:
29.92grams of PbSO4
Explanation:
lead (iV) oxide = PbO2 = Molar mass: 239.2 g/mol
lead (ll) sulfate = PbSO4 = Molar mass: 303.26 g/mol
PbO2 = PbSO4
1:1 ratio
Pb = Lead
Lead has an oxidation number of 4+
O = Oxygen
Oxygen has an oxidation number of 2-
PbO2 + 4H+ + SO4 2- + 2e- = PbSO4(s) + 2H2O
Ok so the above would be the likely complete reaction, though we don't really need this as we already know the ratio is 1:1.
23.6g of PbO2
23.6/239.2 = 0.09866 Moles of PbO2
Since we have a 1:1 ratio we know that the same number of moles of PbSO4 are produced and since we know the molar mass it's simply molar mass multiplied by number of moles.
303.26 x 0.09866 = 29.92grams of PbSO4
