How many grams of the followinf solutes would you need to prepare the indicated volume and concentration of the solutions given?A340mL of a 1.82 M aluminum nitrate solution.B25.0 mL of a 4.26 potassium cyanide solution.
just grease baking sheet really wellit works fine enoughit might stick a bit, but not bad as break cookies.
AFirst, you need to find the mole mass of aluminum nitrate: Al(NO3)3 1 × Al 1 × 26.9815 26.9815 3 × N 3 × 14.0067 42.0201 9 × O 9 × 15.9994 143.9946 -Total 212.9962 g/mol Next, we need to use the following equation to find moles of aluminum nitrate: moles Al(NO3)3 (molarity)(Liters of solution) moles Al(NO3)3 (1.82 M)(0.340 L) moles Al(NO3)3 0.6188 mol Next, we need to multiply the number of moles of aluminum nitrate by its mole mass (see above): (0.6188 mol)(212.9962 g/mol) 131.80204 g or 132 g rounded to three significant figures Answer: We would need about 132 grams of aluminum nitrate to prepare 340mL of a 1.82 M solutionBFirst, you need to find the mole mass of potassium cyanide: KCN 1 × K 1 × 39.098 39.098 1 × C 1 × 12.0111 12.0111 1 × N 1 × 14.0067 14.0067 -Total 65.1158 g/mol Next, we need to use the following equation to find moles of potassium cyanide: moles KCN (molarity)(Liters of solution) moles KCN (4.26 M)(0.0250 L) moles KCN 0.1065 mol Next, we need to multiply the number of moles of potassium cyanide by its mole mass (see above): (0.1065 mol)(65.1158 g/mol) 6.93438327 g or 6.93 g rounded to three significant figures Answer: We would need about 6.93 grams of potassium cyanide to prepare 25.0 mL of a 4.26 M solution.
parchment paper is what i always use
You could probably use foil, especially if it's the heavy barbeque kindDon't forget to rub it with butter or oil or cooking spray.
