I dont even need to know the answer, I would be happy with the steps to get it.
6. Calculate the number of mL of 2.00 M HNO, solution required to react with 216
grams of Ag according to the equation:
3 Ag(s) + 4 HNO3(aq) → 3 AgNO3(aq) + NO(g) + 2 H2O(1)

Volume = 0.33 Liters

Explanation:

For this one, we need to use the equation M= n/ V

n - represents the moles

V- represents the volume

M- represents the concentration

In this problem we know that M= 2M, and we have 216g of Ag, so our first step will be to convert these 216 grams of Ag to moles of Ag, and then convert moles of Ag to moles of HNO using the mole ratios in the equation given.

1 mole of Ag is 108g, so 216g Ag = 0.5 moles Ag

The equation lets us know that 3 moles of Ag compare to 4 moles of HNO3,

so we convert our moles of Ag to moles of HNO3 using this ratio.

0.5 moles Ag * 4 moles HNO3 / 3 moles Ag => 0.66 moles HNO3

Now that we have n and M, we can solve for volume.

2M = 0.66 moles / V

V = 0.66/2 = 0.33 Liters

it's c. elements are placed in rows based on how may valencies electrons have.