Can you plz help me and thankyou so much 4. Can 1.75 L of water dissolve 4.6 moles of Copper Sulfate CusO.?(please show work)Why?/Why not?5. What is the solution concentration formed from 3.6 moles of solute dissolved into 1.3 L of water?(please show work)近

Formula used : Molarity = no. of moles/ volume

4. Can 1.75 L of water dissolve 4.6 moles of Copper Sulfate CuSOa? Why?/Why not? (please show work) 5. What is the solution concentration formed from 3.6 moles of solute dissolved into 1.3 L of water? (please show work)

4. Can 1.75 L of water dissolve 4.6 moles of Copper Sulfate CuSOa? Why?/Why not? (please show work) 5. What is the solution concentration formed from 3.6 moles of solute dissolved into 1.3 L of water? (please show work)

Living By Chemistry: First Edition Textbook

1st Edition

ISBN:9781559539418

Author:Angelica Stacy

Publisher:Angelica Stacy

ChapterU4: Toxins: Stoichiometry, Solution Chemistry, And Acids And Bases

Section: Chapter Questions

Problem SIII7RE

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Addition of water to concentrated sulfuric acid is dangerous because it generates enough heat to boil the water, causing it to spatter out of the container. For this reason, chemists remember to add acid to water. In a dilution experiment, we calculate the amount of the more concentrated solution that must be measured out. If we place this concentrated solution (Figure 4.7) in the volumetric flask first, then dilute with water, we violate the caution Add acid to water. Describe a safer variation on the method shown in Figure 4.7 that allows the quantitative dilution of concentrated sulfuric acid. Figure 4.7 Preparing a dilute solution from a concentrated solution. (a) Draw the concentrated solution into a pipet to a level just above the calibration mark. (b) Allow the liquid to settle down to the calibration line. Touch the tip of the pipet to the side of the container to remove any extra liquid. (c) Transfer this solution to a volumetric flask. Again touch the pipet to the wall of the flask to ensure complete transfer, but do not blow out the pipet because it is calibrated for a small amount of liquid to remain in the tip. (d) Dilute to the mark with solvent. Frequently, especially with concentrated acids, it is best to have some solvent present in the volumetric flask before you add the concentrated sample.
What happens if you add a very small amount of solid salt (NaCl) to each beaker described below? Include a statement comparing the amount of solid eventually found in the beaker with the amount you added: a a beaker containing saturated NaCl solution, b a beaker with unsaturated NaCl solution, c a beaker containing supersaturated NaCl solution. A saturated sodium chloride solution.
11. You have a solution that is 18.2% sodium thiosulfate, Na2S2O3 , by mass. a. What mass of sodium thiosulfate is in 75.0 g of solution? b. How many moles of sodium thiosulfate are in 75.0 g of solution? please show work
2. Calculate the molarity of a solution containing 5 moles of fluorine dissolved in 500L of solution? 3. What is the concentration of a fluorine solution made of 10 moles of solute dissolved in 250 L of water? 4. Find the concentration of 5 moles of solute dissolved in 250 L of water?
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.
46.29 g of KOH is dissolved in 250. mL of water. (Please pay attention to sig figs!) What is the mass percent of KOH in this solution? (use the density of water 1.00g/mL) a. b. What is the concentration, expressed in molal (m = moles solute/kg solvent), of this initial solution? What is the concentration, expressed in molar (M = moles solute/liters solvent), of this initial solution? (assume that the addition of the KOH to the water did not change the C. volume, which is still 250.mL) d. If I add 500.mL of pure water to this 250.mL solution, what is the new concentration expressed in molar (M)? (Careful, what is the new volume after mixing?) e. If, instead, I took the 250. mL solution and evaporated the water until the solution had a volume of 150.mL, what would be the concentration of this this new solution expressed in molar (M)?
PS. This is not graded.
2. What masses of calcium sulfate and water are needed to make 250. g of a 7.5% solution? Show work.
Please breakdown the solution in the same format below. Thank you!
A solution is prepared by dissolving 102 g of NaCl in 924 g water to make a total solution volume of 951 mL. The molar mass of NaCl is 58.44 g/mol. (a) What is the %(m/m) of the solution? %(m/m) (enter the numerical answer only with three sig figs) remember: %(m/m) = g, solute/ total mass solution X 100
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