Chlorine exists primarily as 37cl and 35cl and has an average atomic weight of 35.453. the abundance of cl with atomic weight of 35 is

- Abundance of Cl-35 is 0.7735 or 77.35%.

Solution:- The formula used for calculating the average atomic mass is:

average atomic mass = First isotope mass(abundance) + Second isotope mass(abundance)

Note:- For this formula, we use the decimal abundances, For example if the percent abundance of an isotope is 90% then in decimal it is 0.9.

The sum of abundances of all the isotopes of an element is 1. let's say the abundance of Cl-35 is X then the abundance of Cl-37 would be 1-X. Mass of Cl-35 is 35 and that of Cl-37 is 37 and the average atomic mass is given as 35.453. Let's plug in the values in the formula and solve it for X.

35.453 = 35(X) + 37(1-X)

35.453 = 35X + 37 - 37X

35.453 - 37 = 35X - 37X

-1.547 = -2X

Negative sign is canceled as it's on both sides.

1.547 = 2X

X = 0.7735

So, the abundance of Cl-35 is 0.7735 or 77.35%.

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

m=9

step-by-step explanation: