D.) It depends cuz no yeild is 100%..I mean side reactions also occur in most of the reactions. So mass of the reactant is not equal to the mass of the product. Hope it helps
Answer:
The sediments accumulating on and around mid-ocean ridges are mostly formed from the calcareous and siliceous tests of pelagic organisms. This research is concerned with understanding how the rate of sediment supply varies from place to place due to varied productivity of pelagic organisms, how the sediments accumulate on the complex topography of a mid-ocean ridge, and with using the sediments to study mid-ocean ridge processes such as faulting and volcanism.
Sediment transport and accumulation
When pelagic materials reach the seafloor, they are redistributed by bottom currents and by sedimentary flows. This work studied the form of the accumulation using sediment profiler records collected with a Deep Tow system from the Scripps Institution of Oceanography deployed over the Mid-Atlantic Ridge in the early 1970s. The records showed that both sets of transport processes are important. The shapes of deposits were studied to see to what extent they conform to the diffusion transport model - many deposits have parabolic surfaces, which are the steady state forms expected from the diffusion transport model under boundary conditions of constant input or output flux to basins.
D = m / V
D = 2790 g / 205 mL
D = 13.60 g/mL
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
The characteristic of the Bohr model that would best support his observation is this assumption: "The energy of the electron in an orbit is proportional to its distance from the nucleus. The further the electron is from the nucleus, the more energy it has." The discrete, bright, colored lines might represent the electrons and its distance from the nucleus. The lights are caused by the energy it has.
