The diameter of a chlorine atom is 200. pm. how many chlorine atoms are required to line up end to end to stretch a distance of 1.0 mm? use 'e' for scientific notation.

The number of chlorine atoms needed would simply be the ratio of distance and diameter. But first convert 200 pm to mm:

200 pm = 2 E-7 mm             

So the number of chlorine atoms needed is:

1.0 mm / (2 E-7 mm) = 5,000,000 Chlorine atoms = 5 E6 atoms

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

explanation:

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