Calculate the molarity when 1.24 g agno3 is dissolved to make 125 ml solution. round to two significant digits.

Molarity=0.0058M

Explanation:

Given the weight of AgN is(w) 1.24g

We know that the molecular weight of AgN is(W) 169.87g/mol.

Given the volume of the solution is(V) 125ml.

Here solute is silver nitrate and the solvent is water.

We know that

We know that number of moles=weight/molecular weight

=  

volume of the solution(V)=125/1000=0.125L

Now

Molarity=n/V

M=     =0.00584M

the answer for sodium is 0.186

the answer for rubidium is 0.251

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.