You are given a test tube containing 10 mL of a solution with 8.4 × 107 cells/mL. You are to produce a solution that contains less than 100 cells/mL. What dilutions must you perform in order to arrive at the desired result?

(1). 1 ml of original solution to 99 ml of water = 8.4 ×

(2). 1 ml of second solution to 99 ml of water = 8.4 ×

(3). 1 ml of third solution to 99 ml of water = 84  

Explanation:

Volume of solution = 10 ml

We should perform the following dilutions.

(1). 1 ml of original solution to 99 ml of water = 8.4 ×

(2). 1 ml of second solution to 99 ml of water = 8.4 ×

(3). 1 ml of third solution to 99 ml of water = 84  

These are the three dilutions we have to perform.

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