Percent of silver by mass in an unknown sample
Noah--

You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:

82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)

0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)

2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass

I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.

And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.