he following five beakers, each containing a solution of sodium chloride (NaClNaCl, also known as table salt), were found on a lab shelf:   Beaker Contents 1 200. mLmL of 1.50 MM NaClNaCl solution 2 100. mLmL of 3.00 MM NaClNaCl solution 3 150. mLmL of solution containing 25.5 gg of NaClNaCl 4 100. mLmL of solution containing 25.5 gg of NaClNaCl 5 300. mLmL of solution containing 0.450  molmol NaClNaCl   Arrange the solutions in order of decreasing concentration. Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them.

O Chemical Reactions = Calculating molarity using solute moles 3/5 Bisho A chemist prepares a solution of barium acetate (Ba (CH3CO2)2) by measuring out 28.2 μmol of barium acetate into a 50. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's barium acetate solution. Round your answer to 2 significant digits. mol X 15

4.87 mL of a dark solution with a dye concentration of 15.86 %m/v is diluted to a total  volume of 146. mL. what is the new concentration of solution in %m/v ?

What is the molarity of a solution in which 4.00 mol of potassium chloride (molar mass 74.55 g/mol) is dissolved in 3.00 L of solution? O 2.67 M 4.00 M 6.00 M O 1.33 M MacBook Air D00 20 F3 F10 esc F4 F5 F7 F8 F9 F6 F1 F2 @ #3 2$ % 2 3 6 8 Q W E R Y ーの

Equal volumes of Solutions 2 and 3 were mixed in the lab. Suggest a measurement that could be made in the lab to provide evidence that a chemical reaction occurred when Solutions 2 and 3 were mixed.

solution. You have 2.5ml of a 1mg/mL NaCl solution. You need to make 5 mL of a 0.25mg/mL HAVE (Stock solution) C₁ 1mg/mL NaCl V₁ = ? mL Please note: NEED (New solution) C₂ = 0.25mg/mL V₂ = 5mL • The needed/ new solution is more dilute (less concentrated) than the stock solution. • It does not matter how much (volume) of stock solution one has, what matters is how much is needed to make a new solution of given concentration and volume. • The units of volume and concentration on both sides of the equation must be the same. C*Vi C2 * V2 (1mg/mL) (V1mL) (0.25mg/mL) * (5mL) V1 mL (0.25mg/ml) * (5mL) = 1mg/mL 1.25 mg 1 mg/ml V₁ = 1.25mL Even though you had 2.5mL of stock solution, only 1.25mL of it is needed to make the diluted solution. • The final volume of the new solution is 5mL. You will need to add (V1-V2 = 5ml - 1.25mL) 3.75mL of diluent to the 1.25mL stock solution.

9.8 g of H2SO4 (molar mass 98 g/mol) is present in 2.0 L of solution. The molarity of the solution is: 0.010 M 0.10 M 0.050 M 0.20 M MacBook Air ►► esc 20 F1 F3 F4 F6 F7 F8 F9 F10 F2 #3 %2$ % & 2 3 5 7 8 9.

Complete the table below by determining the missing quantity in each sample. All solutions are aqueous. Any quantity that is not applicable to a given solution is marked NA. Show your works

A student wants 83.0 mL of a 11.5% sugar water solution. They only have a 17.0% stock solution. How many milliliters of the 17.0% stock solution should they use to make the 11.5% sugar water solution?     14.0 mL     56.1 mL     21.1 mL     62.1 mL     54.8 mL

10. A 4 M solution of NaOH means a. 4 moles of NaOH in 1 L b. 4 grams of NaOH in 1L c. 4 moles of NaOH in 100 mls 11. What is a solvent? a. The portion of a mixture present in the largest quantity b. The portion of a mixture present in the smaller quantity 12. If we dissolve 116 grams of NaCl in one Liter of water, what is the Molarity of the solution? c. 58 M d. 1 M e. 2 M 13. We have 12 M HCl and need 4 L of 6 M HCl. How many Liters of 12 M HCl should we use? a. 3 L b. 4 L c. 2 L