If you’re asking to balance the equation then:
Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)
Just remember: the equations at the end is Cl not C12
Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
It’s hard which class are u
Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:
54.3 mm = 5.43 cm
Now, we determine the number of gold atoms that will be present in this:
5.43 / 1 x 10⁻⁹
There will be 5.43 x 10⁹ atoms
We now determine the number of moles this is by:
one mole = 6.02 x 10²³ atoms
Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles
The molar mass of gold is 197 g/mol
The mass is 9.01 x 10⁻¹⁵ * 197
The mass of the strand is 1.76 x 10⁻¹² grams
