D. CuCl2 copper(2)chloride
Answer:
The concentration of the most dilute solution is 0.016M.
Explanation:
First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:
![[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M](https://tex.z-dn.net/?f=%5BNa_%7B2%7DSO_%7B4%7D%5D%3D%5Cfrac%7Bmoles%28Na_%7B2%7DSO_%7B4%7D%29%7D%7Bliters%28solution%29%7D%20%3D%5Cfrac%7Bmass%28%28Na_%7B2%7DSO_%7B4%7D%29%29%7D%7Bmolarmass%28moles%28Na_%7B2%7DSO_%7B4%7D%29%20%5Ctimes%200.100L%29%7D%20%3D%5Cfrac%7B2.5316g%7D%7B142g%2Fmol%5Ctimes%200.100L%20%7D%20%3D0.178M)
<u>First dilution</u>
We can use the dilution rule:
C₁ x V₁ = C₂ x V₂
where
Ci are the concentrations
Vi are the volumes
1 and 2 refer to initial and final state, respectively.
In the first dilution,
C₁ = 0.178 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
<u>Second dilution</u>
C₁ = 0.053 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
Answer:
All rocks have same thermal conductivity.
Explanation:
Rocks are heated with the sunlight exposure. Different types of rocks have different thermal conductivity. The heat energy of the rocks is measure in Joules. The size and thermal conductivity of the rocks is important property which determines their heating capacity.
The net ionic equation is
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Molecular equation
:</em>
Cu(s) + 4HNO₃(aq) ⟶ Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Ionic equation:
</em>
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Net ionic equation
</em>
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Note</em>: The net ionic equation is <em>the same as </em>the ionic equation because there are <em>no common ions</em> to cancel on opposite sides of the arrow.
