If you have 112 g of Ni and 112 g of AgNO3, how many atoms of silver would be produced?

3.97× 10²³ atoms of Ag  

Explanation:

Given data:

Mass of Ni = 112 g

Mass of AgNO₃ = 112 g

Atoms of silver produced = ?

Solution:

Chemical equation:

Ni  + 2AgNO₃    →     2Ag + Ni(NO₃)₂

Number of moles of Ni:

Number of moles = mass/ molar mass

Number of moles = 112 g/58.7 g/mol

Number of moles = 2 mol

Number of moles of AgNO₃:

Number of moles = mass/ molar mass

Number of moles = 112 g/169.87 g/mol

Number of moles = 0.66 mol

Now we will compare the moles of Ag with both reactants.

                             AgNO₃          :            Ag

                                 2                :            2

                              0.66             :           0.66

                                 Ni               :           Ag

                                  1                 :            2

                                  2                ;          2×2= 4 mol

Number of moles of Ag produce by  AgNO₃  are less so it will limiting reactant and limit the yield of Ag.

I mole = 6.022 × 10²³ atoms

0.66 mol ×  6.022 × 10²³ atoms / 1 mol

3.97× 10²³ atoms of Ag                              

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3