Answer:
NaNO₃
Explanation:
A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.
From the choices given precipitation reaction will occur between;
- Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
- Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
- FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)
Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.
From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.
An atom of element has there subatomic particles namely, proton, electron and neutron. Here, for a neutral atom, number of proton is equal to number of electron (this is not in the case of ions), this is equal to atomic number of an atom. In an atom, nucleus contains protons and neutrons which is responsible for mass of the atom and electrons move around nucleus in fixed orbits. Thus, atomic mass of an atom is equal to sum of number of protons and neutrons.
Option (b): Proton is the particle in nucleus of an atom, whose total number is equal to atomic number of that atom.
(4) Option (b): Atoms of same element have same atomic number because mass number can be different for different isotopes of atom. Since, atomic number is equal to number of protons, thus, number of protons are same for all atoms of the same element.
(5) Option (d): Isotopes are defined as atoms of same element with same atomic number but different mass number. Thus, correct option is (d) mass numbers.
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
