How many grams of AgNO3 (silver nitrate) are contained in 425 mL of a .15 M solution of AgNO3?

mols = M x L = ?

mol = grams/molar mass. You know molar mass and mols, solve for grams.

i don't understand...

You can't understand if you don't try. You could help by explaining exactly what you don't understand instead of leaving me hanging out there with no idea of which direction to go to help.

mols = M x L. You have M = 0.15M in the problem. You have 425 mL (0.425L) volume in the problem. 0.425L x 0.15M = ? mols AgNO3. You can multiply that on your calculator.
Then mols AgNO3 = grams AgNO3/molar mass AgNO3. You have mols from the previous line. You know molar mass AgNO3. Substitute those two values and solve for grams AgNO3. That's the answer.

To calculate the number of grams of AgNO3 (silver nitrate) in a given volume of a solution, you can use the formula:

Number of grams = Concentration (M) * Volume (L) * Molar mass

First, let's convert the given volume from milliliters (mL) to liters (L):

425 mL ÷ 1000 = 0.425 L

Next, we need to determine the molar mass of AgNO3. The molar mass can be calculated by summing the atomic masses of each element in the compound, which are as follows:

Ag (silver) = 107.87 g/mol
N (nitrogen) = 14.01 g/mol
O (oxygen) = 16.00 g/mol (x3 since there are three oxygen atoms in AgNO3)

Molar mass of AgNO3 = (1 * 107.87) + (1 * 14.01) + (3 * 16.00) = 169.87 g/mol

Now, we can substitute the values into the formula:

Number of grams = 0.15 M * 0.425 L * 169.87 g/mol

Multiply the concentration and volume:

Number of grams = 0.06375 * 169.87 g/mol

Finally, calculate the product:

Number of grams ≈ 10.845 grams

Therefore, there are approximately 10.845 grams of AgNO3 in 425 mL of a 0.15 M solution of AgNO3.