7.4x10^23 = molecules of silver nitrate sample
6.022x10^23 number of molecules per mole (Avogadro's number)
Divide molecules of AgNO3 by # of molecules per mol
7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)
(I leave off the x10^23 because they both will divide out)
Use your periodic table to find the molar weight of silver nitrate.
107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3
Now multiply your moles of AgNO3 with your molar weight of AgNO3
1.229mol x 169.868g/mol = 208.767g AgNO3
That's the answer on that picture
Answer:
1. The metal atom/ion in these compounds are Ni and Ni2+ respectively.
2. The electrons from s oribital will jump to d orbital and so I expect CO to donate electron pairs in 4p and 4s orbitals and form sp3 hybridisation.
Answer:
<h2>18 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

volume = final volume of water - initial volume of water
volume = 62 - 52 = 10 mL
From the question we have

We have the final answer as
<h3>18 g/mL</h3>
Hope this helps you