Acertain element x has four isotopes. 4.350% of x has a mass of 49.94605 amu. 83.79% of x has a mass of 51.94051 amu. 9.500% of x has a mass of 52.94065 amu. 2.360% of x has a mass of 53.93888 amu. what is the average atomic mass of element x? express your answer numerically to four significant figures. view available hint(s)

Average atomic mass of an element is the sum of atomic mass of its isotopes multiplied by their respective percentage abundance.

There are four isotopes of element X:

1. 4.350 % with mass 49.94605 amu

2. 83.79% with mass 51.94051 amu

3. 9.5% with mass 52.94065 amu

4. 2.360% with mass 53.93888 amu

Thus, average atomic mass will be:

Or,

Therefore, average atomic mass of element X will be 51.99 amu (four significant figures)

52.00 amu.

Explanation:

Mass of isotope 1 = 49.94605 amu

% abundance of isotope 1 = 4.350% =

Mass of isotope 2 =  51.94051 amu

% abundance of isotope 2 = 83.79 % =

Mass of isotope 3 =  52.94065 amu

% abundance of isotope 3  = 9.500 % =

Mass of isotope 4 = 53.93888 amu

% abundance of isotope 4 = 2.360 % =

Formula used for average atomic mass of an element :

Therefore, the average atomic mass of an element is 52.00 amu.

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

d

explanation: