Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.
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.
the answer is A. Does temperature affect the growth of rose bushes because remember their asking u what question can you answer doing the investigation.
Molarity is defined as the number of moles of solute in 1 L of solution
mass fo KCl in the solution is - 5.0 g
number of moles of KCl - 5.0 g/ 74.5 g/mol = 0.067 mol
number of moles of KCl in 100 mL - 0.067 mol
therefore number of KCl moles in 1 L - 0.067 / 100 mL x 1000 mL = 0.67 M
molarity of KCl is 0.67 M
