Answer:
d. N
Explanation:
Chemical equation:
Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + KNO₃(aq)
Balanced Chemical equation:
Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + 2KNO₃(aq)
Ionic equation:
Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Net ionic equation:
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
The NO₃⁻(aq) and K⁺(aq)are spectator ions that's why these are not written in net ionic equation. The PbSO₄ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Phosphorus has a larger atomic radius than Sulfur
Answer: -
The approximate number of atoms in a bacterium is 10¹¹
Explanation: -
We are given the mass of a bacterium is 10⁻¹⁵ kg.
We are told that the mass of a hydrogen atom is 10⁻²⁷ kg.
Finally we learn that the average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.
Mass of an atom of bacterium = 10 x mass of hydrogen atom
= 10 x 10⁻²⁷ kg.
= 10⁻²⁶ kg.
Thus the number of atoms in a bacterium =
=
= 10¹¹
Answer:
Krypton is a colorless, odorless gas. It has a boiling point of -152.9°C (-243.2°F) and a density of 3.64 grams per liter. That makes krypton about 2.8 times as dense as air.
First, we need to find the
limiting reactant. <span>
149 g Al x 1 mole/27g Al = 5.52 mol Al
601 g Fe2O3 x 1 mole/159.6g Fe2O3 = 3.77 mol Fe2O3
<span>Al is the limiting reactant since 3.77 mol Fe2O3 would
require 3.77*2 = 7.54 mol Al but the given is less.
<span>mol Al2O3 = 5.52 mol Al * 1 mol Al2O3 / 2 mol Al
= 2.76 mol</span>
mass Al2O3 = 2.76 mol x 102g/1 mole AlO3 = 281.52 g Al2O3
will be formed. </span></span>
<span>
Since 3.77 mole of Fe2O3 is present, but 5.52 mole Al is the
limiting reactant, then only 5.52/2 mole Fe2O3 can be used. This leaves an
excess of:</span>
Fe2O3 excess = 3.77 – 5.52/2
= 1.01 <span>
<span>1.01 mole Fe2O3 x 103.8g/1 mole Fe2O3 = 104.84 g Fe2O3 </span></span>