How many mL of 3.00 M CaCl2 solution would have to be diluted with water to produce 500.mL of 0.500 M solution? a. 167 b. 83.3 c. 60.2 d. 30.0 e. no correct response
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6. How many mL of 3.00 M CaCl2 solution would have to be diluted with water to produce 500.mL of 0.500 M solution?
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- How many mL of 0.668 M HBr are needed to dissolve 7.69 g of MgCO,? 2HB1(aq) + MgCO3(s) MgBrz(aq) + H50(1) + CO2(g) mL 9 more group attempts remaining Retry Entire Group Submit Answer hp ins prt sc dele f12 f11 トト f8 f9 f10 f6 17 L. 80What volume (in mL) of 0.2300 M HBr is required to neutralize 50.00 mL of 0.7000 M KOH? +/- P %#3 % & 3 4. 6. 7. R. T. command101 Bb UTF-8"Lab%205-%20PI x Bb UTF-8"Lab%202-%20st Bb UTF-8"Lab%201-%20SCX I https://mssu.yellowdig x Question 4 of 34 If you have 155 mL solution of a 0.762M FeCl solution, how many moles of FeCl, are contained in this sample? mlsetuton ADD FACTOR DELETE ANSWER RESET *( ) 1 1000 162.2 10 35.45 6.60 55.85 0.01 12.6 0.203 100 0.118 0.1 0.001 155 4.92 0.762 19.2 11.8 118
- b. 8.0 moles of Na2 SO4 in 2.50 L of solution MNa+ M M80,- MWhich of the following is the least soluble in water? Select one: O a. NaCI (sodium chloride) O b. NH4NO3 (ammonium nitrate) O c. Na2CO3 (sodium carbonate) O d. CaSO4 (calcium sulfate) ge 1 Chapters 1 and 2 Written Response Jump to. Accinnmont 58.pdf Chap6 SF10 (1) pdf Files (1).zipBenzalkonium Chloride (E=0.16) =1.5g Distilled water qs 0.9% NaCl solution ad =100 mL Make isoton. Sol. Apply on affected area p.r.n. 1. How many mL of distilled water must be added to make the Benzalkonium chloride preparation isotonic? 2. How many mL of 0.9% NaCl must be added to make a 100 mL solution?
- What is the gram solubility (in g/L) of Ag,So, in water, given its Ksp 1.40 x 105 and its Molar Mass is 311.80 g/mol. Leave answer in 3 significant figures (e.g. 30.2).1. How many moles of HNO3 are needed to prepare 5.0 L of a 2.0 M solution of HNO3? а. 2.5 b. 5 С. 10 d. 203. Which of the following statement/s is/are correct? 1. 6.00 N H2CO3 is equivalent to 3.00M H2CO3. II. 5.29 % (w/v) H2SO4 is equivalent to 0.539 N H2SO4 II. 1.15 N Mg(OH)2 is equivalent to 16.3 g Mg(OH)2 in 500 mL solution. (MH2S04 = 98.086 g/mol; MH2C03 = 62.026 g/mol; MMg(OH)2 = 58.326 g/mol) %3D
- (₁ 2. X 70 R R t X 3. R-= X excessive 1) Na NH₂ 2 2) H₂O D Nạnh 2 2) H₂0 excessive HX X x+ KA. Standardlzatlon of thlosulfate solutlon Trial 1 Trial 2 Trial 3 Initial volume of thiosulfate (mL) 0.00 20.05 2.3 Final volume of thiosulfate (mL) 20.05 40.05 22.15 Volume of thiosulfate used (mL) Molarity of thiosulfate Average Molarity of thiosulfate Calculate the molarity of the thiosulfate solution using the formula: 6 mmol S20,²- 1 mmol 10, Ms,0,2- 0.004167 M 10,- 20.00 ml Vol S20,²- used (mL)Ans. 17.15 mg. 54) A 1.000-g sample of a mixture which contains only NaCl and KCl gare a precipitate of AgCl which weighed 2.000 g. What are the percentages of Na and K in the mixture? ins 10 g Ans. 44.77% K; 5.75% Na. 1:J 55. I- can be sernarated from othor