Answer:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Explanation:
Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.
3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)
The complete ionic equation includes all the ions and insoluble species.
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Answer:
2.28 m
Explanation:
Use the relationship
λ=vf
to solve for wavelength λ. Substituting the known quantities yields:
λ=vf
λ=2.99 x 108 m/s1.31 x 108 Hz
λ=2.28 m
Answer:
productivity and water depth
Explanation:
The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.
The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.
Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.
The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
Answer:
4( atoms of the same element always have different atomic weights)
Explanation:
the number of protons in the atoms of the same type can be different especially in isotopes
