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Anni [7]
3 years ago
7

Let’s say that you have a solvatochromic compound that appears red in a solvent. You dissolve the compound in another solvent an

d it changes to an orange color. Would this be considered a hypsochromic or bathochromic shift? Would the solvent be more polar or less than the first solvent? Explain
Chemistry
1 answer:
EastWind [94]3 years ago
7 0

Answer:

Hypsochromic compound, More polar solvent

Explanation:

Hypsochromic shift refers to the shift of solution colour to blue side of the visible spectrum (blueshift) with increasing polarity of the solvent. In our case, the solution changes to orange colour from red when solvent is changed. This means that the emission spectrum of the solution underwent blueshift. (As orange colour is on the 'blue' side for red colour.) So this is a hypsochromic shift, and the new solvent is more polar that the previous one, as it caused hypsochromic shift.

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lozanna [386]

Answer:

The answer to your question is letter A. 2

Explanation:

Data

Octane = C₈H₁₈

Oxygen = O₂

Carbon dioxide = CO₂

Water = H₂O

Reaction

                         C₈H₁₈  +   O₂   ⇒   CO₂   +   H₂O

                  Reactants         Elements          Products

                          8                     C                         1

                         18                     H                        2

                          2                     O                        3

This reaction is unbalanced

                          C₈H₁₈  +  25/2O₂   ⇒   8CO₂   +   9H₂O

                  Reactants         Elements          Products

                          8                     C                         8

                         18                     H                        18

                         25                     O                       25

Multiply the reaction by 2

                    2C₈H₁₈  +   25O₂   ⇒   16CO₂   +   18H₂O

                  Reactants         Elements          Products

                         16                     C                        16

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Now, the reaction is balanced

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