Barrier islands typically have sand in the beach zone and dune field, and mud in the back-barrier. Overwash deposits sand in the back-barrier.
Barrier islands form in three ways. They can form from spits, from drowned dune ridges or from sand bars. Longshore drift is the movement of sand parallel to the shore caused by the angle of the waves breaking on the beach. ... When a storm such as a hurricane digs an inlet through the spit a barrier island is formed.
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.
Concentration is defined as the number of moles of a solute present in the specific volume of a solution.
According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.
M₁V₁=M₂V₂
Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml
Rearrange the formula for M₂
M₂=(M₁V₁/V₂)
Plug all the values in the formula
M₂=(1.0M×14 ml/25 ml)
M₂=14 M/25
M₂=0.56 M
Therefore, the concentration of a dextrose solution after the dilution is 0.56M.
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