Why do we use acetic anhydride for this reaction? List two other possible sources of the acyl group and why they would be a better/worse choice for this reaction, mechanistically, practically, and by the principles of green chemistry.

Read these sentences from the text. near the equator, the tropics receive the most rain on a consistent basis. as a result, the fresh water falling into the ocean decrease the salinity of the surface water in that region. [. .] . . as the salt content of sea water increases, so does its density. what can you infer about how rain affects the density of surface water near the equator?

In pea plants, the allele for tallness (t) is dominant to the allele for shortness (t). in the cross between a tall pea plant and a short pea plant shown below, what is the probability that the resulting offspring will be tall? whats the percent

Ineed someone to see if my answers are correct! if any are wrong let me know what the correct answers would be and how to get that answer! 1. how many moles of sodium chloride are in 28 grams od nacl? a. 265 mole naclb. 856 mole naclc. 479 mole of nacld. 1.2 mole nacl < my choice2. 734 grams of lithium sulfate (li2so4) are dissolved to make 2500 ml of solution what is rhe molaratiy? a. 2.67 mb. 4.56 mc. 3.89 m < my choiced. 1.78 m3. how many grams of cacl2 would be dissolved in 3.0 l of a 0.50 m solution of cacl2? a. 250 g cacl2 b. 166.5 g cacl2c. 113.65 g cacl2d. 98 g cacl2 < my choice4. suppose you had 58.44 g of nacl and you dissolved it in exactly 2.00 liters. the molarity if the solution would be 0.5 mtrue < my choicefalse 5. i would need 22g of naoh to make a 3.0 m solution using 250 ml of solvent.true < my choicefalse6. identify the solute: you have a .0195 m solution made from using 6.5 g of solute and 3 l of solvent. identify the solute by solving for molar weight.a. the solute is nacl because the molar weight is 58.43 g/mol < my choiceb. the solute is h2so4 because the molar weight is 98.06 g/molc. the solute is cacl2 because the molar weight is 111.11 g/mol