Answer:
b. HCOOH/ NaHCOO.
Explanation:
A buffer system may be formed in one of two forms:
- A weak acid with its conjugate base.
- A weak base with its conjugate acid.
Chose the pairs below that you could use to make a buffered solution.
a. HCI/NaOH. NO. HCl is a strong acid and NaOH is a strong base.
b. HCOOH/ NaHCOO. YES. HCOOH is a weak acid and HCOO⁻ (coming from NaHCOO) is its conjugate base.
c. HNO₂/H₂SO₃. NO. Both are acids and they are unrelated to each other.
d. NaNO₃/ HNO₃. NO. HNO₃ is a strong acid.
Elements are substances that contain only 1 kind of atom.
a) The E might belong to group 13.
As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.
Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.
The Group 13 is the boron group which has the following elements:
- Boron
- Aluminium
- Gallium
- Indium
- Thallium
All these elements have the valency of +3.
To know more about Valency, refer to this link:
brainly.com/question/12717954
#SPJ4
N₂ : limiting reactant
H₂ : excess reactant
<h3>Further e
xplanation</h3>
Given
mass of N₂ = 100 g
mass of H₂ = 100 g
Required
Limiting reactant
Excess reactant
Solution
Reaction
<em>N₂+3H₂⇒2NH₃</em>
mol N₂(MW=28 g/mol) :
mol H₂(MW= 2 g/mol) :
A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants
From the equation, mol ratio N₂ : H₂ = 1 : 3, so :
N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant
