OH- is common to bases.
Explanation:
The base is a is an ionic compounds which when placed in aqueous solution dissociates in to a cation and an anion OH-.
The presence of OH- in the solution shows that the solution is basic or alkaline.
From Bronsted and Lowry concept base is a molecule that accepts a proton for example in NaOH, Na is a proton donor and OH is the proton acceptor.
A base accepts hydrogen ion and the concentration of OH is always higher in base.
There is a presence of conjugate acid and conjugate base in the Bronsted and Lowry acid and base.
Conjugate acid is one which is formed when a base gained a proton.
Conjugate base is one which is formed when an acid looses a proton.
And from the Arrhenius base Theory, the base is one that dissociates in to water as OH-.
Answer:
Correct answer is option (3) .
Explanation:
Hope it helpful....
A scientific theory is a well tested explanation of a natural phenomenon.
When two gases of a chemical reaction are at the same temperature, pressure and molar volume, then the stoichiometric ratio of the gases would be 1 is to 1. Molar volume is the volume of the gas per mole of the gas. Having the same conditions for both gases would mean that they are present with the same number of moles.
Answer:

Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant

3. Calculate the moles of Pb from each reactant

4. Calculate the mass of Pb
