How much agarose powder would you need to use (include units! ) if you only needed 40 ml of agarose gel and the package recipe states ‘use 6 grams of agarose powder dissolved in 500 ml of water’?

The rule of three is a way to solve problems of proportionality between three known values and one unknown.  This is an operation of finding the fourth term of a proportion knowing the other three.

We can use the rule of three to solve this problem in this way,

500 mL →  6 grams

40 mL  →     x

This is the same as saying: if in 500 mL there should be 6 grams of agarose powder then in 40 mL how many grams should there be?

So,

→ x = 0.48 g = 480 mg

Then, you need to dissolved 480 mg of agarose powder in 40 mL of water to meet the recipe.

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