<u>Answer:</u> The products of the given chemical equation are 
<u>Explanation:</u>
Protonation equation is defined as the equation in which protons get added in the substance.
The chemical equation for the protonation of carbonate ion in the presence of water follows:
By Stoichiometry of the reaction:
1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion
Hence, the products of the given chemical equation are
Answer:
The mass of the reactants compared with the mass of the products should be the same if the reactants are in stoichiometric amounts.
Explanation:
In this question, they ask about chemical reactions and the comparison of the mass of reactants and products. Firstly, it is necessary to introduce the mass conservation principle.
Mass conservation principle mentions that in a chemical reaction, the total mass of reactants is equal to the total mass of products (if the reaction is fully developed). It means mass is not created or destroyed, only transforms from reactants to products.
For example, the mass of sodium plus the mass of chlorine that reactswith the sodium equals the mass of the product sodium chloride.Because atoms are only rearranged in a chemical reaction, there mustbe the same number of sodium atoms and chlorine atoms in both thereactants and products.
Finally, we can conclude that The mass of the reactants compared with the mass of the products should be the same if the reactants are in stoichiometric amounts.
Answer:
0, l is n-1 always, ml is l to -l
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
