A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

Likely (indium.)

Explanation:

Number of atoms: .

Dividing, , the number of atoms by the Avogadro constant, , would give the number of moles of atoms in this sample:

.

The mass of that many atom is . Estimate the average mass of one mole of atoms in this sample:

.

The average mass of one mole of atoms of an element ( in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass . Indium, , is the closest match.

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

particle   p is an electron and particle q is a neutron