For a particular lab experiment, you need 150 mL of a 2.50 M solution of ammonium nitrate. How many grams of NH4NO3 will you need for the solution?

12.0 M=3.00 mol/xx=0.250 L. This calculates the volume in liters. 0.250L x 1000 mL / L= 250. mL (note use of explicit decimal point to create three sig figs) (0.250 molL¯1) (0.100 L) = x / 74.0918 g mol¯1 x = (0.250 mol L¯1) (0.100 L) (74.0918 g mol¯1) x = 1.85 g (to three sig figs) (x) (0.0500 L) = 20.0 g / 97.9937 g mol¯1 (x) (0.0500 L) = 0.204094753 mol x = 4.08 M(0.500 mol L¯1) (2.50 L) = x / 74.551 g mol¯1 x = 93.2 g(x) (0.2500 L) = 12.0 g / 39.9969 g mol¯1 note use of L in set up, not mL x = 1.20 M x = 4.67 mol / 2.04 L x = 0.629 mol / 1.500 L (x) (10.00 L) = 4.783 g / 106.0 g mol¯1 x) (0.250 L) (= 0.897 g / 96.09 g mol¯1 e) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯1 x = (2.00 mol L¯1) (2.35 L) x = (0.415 mol L¯1) (0.01600 L) x (0.500 mol L¯1) (3.00 L) d) x = (3.76 mol L¯1) (6.20 L) (0.00300 mol L¯1)(0.289L)=x/187.85gmol¯1(5.90mol L¯1)(0.01600L)=x/331.2g mol¯1(2.75molL¯1)(0.508L)=x/41.99gmol¯1(3.76mol L¯1(6.20L)=x/61.98gmol¯1(1.00molL¯1)(0.500L)=x/74.55gmol¯1(3.50mol L¯1)(4.35 L)=x/110.99gmol¯1

for points guys

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong