Michael is studying the factors that affect corrosion of metals. He wants to test the effect that different concentrations of an acid will have on the corrosion of a metal. As he sets up and completes his experiment, which of the following experimental designs would be most likely to help him answer his question? A. Allow several metal samples to soak in different amounts of acid of a single concentration and then measure the amount of corrosion on each. B. Prepare solutions of different acid concentrations, measure 50 milliliters of each into different beakers, and place metal samples of different types but of the same mass into the beakers. C. After allowing several small metal sheets to be exposed to different acid concentrations, measure the surface area of each that is corroded. D. After allowing a small metal sheet to be exposed to one acid concentration, place it in different acid concentrations, then measure the surface area that is corroded.

B. Prepare solutions of different acid concentrations, measure 50 milliliters of each into different beakers, and place metal samples of different types but of the same mass into the beakers.

Explanation:

Since Michael is studying the factors that affect corrosion of metals, and wants to test the effect that different concentrations of an acid will have on the corrosion of a metal, using different acid concentrations of the same volume on different metal samples of the same size would give a much better than all the other options because, a standard volume of acid is used and also a standard mass of metal is used. So, the effect can be measured quantitatively since we have the same quantity of sample.

So, option B is the answer

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

c. 75.7

i think that is the right