In a chemical process, three bottles of a standard fluid are emptied into a larger container. A study of the individual bottles shows that the mean value of the contents is 13 ounces and the standard deviation is 0.07 ounces. If three bottles form an i. i.d. sample,
(a) Find the expected value and the standard deviation of the volume of liquid emptied into the larger container.
(b) If the content of the individual bottles is normally distributed, what is the probability that the volume of liquid emptied into the larger container will be in excess of 39.1 ounces?

Identify one disadvantage to each of the following models of electron configuration: dot structures arrow and line diagrams written electron configurations type in your answer below.

Identify the missing numbers below to show the result of multiplying the numbers (1.6 × 10-19)(5.0 × 106) = c × 10d

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