Please help make corrections thank you Select the keyword or phrase that will best complete each sentence.Key terms:Toluene reacts faster than benzene in all substitution reactions. Thus, its electron-donating CH, group activates the benzene ring to electrophilic attack.acidactivatesA resonance effect is electron rich when resonance structures place a negativecharge on carbons of the benzene ring.basedeactivatesInductive effects stem from the polarizability of the atoms in the substituent andthe electronegativity of the substituent groups.donatingIn a Friedel-Crafts alkylation, benzene is treated with an alkyl halide and a Lewisacid (AICI,) to form an alkyl benzene.donationelectronegativityWhen a neutral O or N atom is bonded directly to a benzene ring, the donatingeffect dominates and the net effect is electron donationo.inductivepolarizabilityNitrobenzene reacts more slowly than benzene in all substitution reactions. Thus,its electron-withdrawing NO, group deactivates the benzene ring.poorresonanceA resonance effect is electron withdrawing when resonance structures place apositive charge on carbons of the benzene ring.richwithdrawalWhen a halogen is bonded to a benzene ring, the inductive effect dominates andthe net effect is electron withdrawingwithdrawingAn alkyl group is an electron-donatingo group and an alkyl benzene is moreelectron richo than benzene.

Rich -------->donating Polarizability -------> Electronegativity Electronegativity------>…

Answered: penzene is treated with a penzene. =…

penzene is treated with a penzene. = bonded directly to a ber ffect is electron donation wly than benzene in all su group deactivateso the withdrawing when reso

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A student weighed out 0.150 g of protein powder and dissolved it in 100 mL of water (Solution 1). The student then diluted this solution by transferring 1 mL into a 25 mL flask and diluting with water (Solution 2). Finally, 1 mL of that solution was transferred to a test tube and combined with 4 mL Bradford reagent. The absorbance of the solution in the test tube was 0.11. Assuming that the best fit linear line of the standard curve was y = 0.04144 x + 0.01521 (μ g mL), calculate the percent protein by mass in the original protein powder.
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