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- View Assessment ← C Pri Ter Acc X 13 minutes remaining Question 1 blackboard.matc.edu/ultra/courses/_525703_1/outline/assessment/_1556 Blank 1 X Blank 2 Blank 3 All answers must have the correct units. Watch your significant digits! A 55.0 g sample of sugar (C12H22011) is dissolved in 170. g of water. a. What is the mass of the solute? Blank 1 g b. What is the mass of the solvent? Blank 2 g c. What is the mass of the solution? Blank 3 g d. What is the percentage by mass of this solution? (% m/m) Blank 4 % Blank 4 Add your answer Add your answer Add your answer Add your answer Courses Last saved 12:59:40 PM X Questions Filter (6) ▼ + K First < Previous NextConstants I Peric 1.00 g KCl in 95.0 mL of solution Express your answer with the appropriate units. µA M = Value Units Submit Request Answer Part B 1.00 g Na2CrO4 in 95.0 mL of solution Express your answer with the appropriate units. HẢ M = Value Units C.1. (answer the questions in the tan boxes in the tables below each question) A student conducted an investigation to determine the effect of water temperature on the amount of sugar that dissolves in a beaker of water. Identify components for trial 1 of this investigation. Beaker Number 1 2 3 4 Amount of Water (mL) 100 100 100 100 Amount of Water (mL) Trial 1 Temperature of Temperature of Amount of Sugar Sugar (°C) Water (°C) Dissolved (g) 20 20 20 20 Temperature of Sugar 5 10 15 20 185 189 194 204 Temperature of Water Terms Variable Constant Amount of Sugar Dissolved
- 14406e994a5937753d2#10001 Constants I F plutions ved solute? 25.0 g solute in 0.325 M NaF Express your answer with the appropriate units. HẢ ? V = Value Units Submit Request Answer Part B 25.0 g solute in 0.325 M CdCl2 Express your answer with the appropriate units. HẢ V = Value Units Submit Request AnswerSolute amount Solution volume Solution Concentration (moles) (Liters) (Molarity) 1.000 0.500 0.500 0.500 0.600 0.833 0.500 0.700 0.714 0.500 0.800 0.625 0.500 0.900 0.556 0.500 1.000 0.500 Show the calculation process to find the concentration of the solution when the solution volume changes from 0.500 L to 0.600 L.Principal ingredient of an antacid is Al(OH)3 (76.98). The concentration of 0.00400 M Al(OH)3 solution in terms of normality is a. 4.0 x 10-3 N b. 1.3 x 10-3 N c. 1.2 x 10-2 N d. 8.0 x 10-3 N
- How many moles of solute in 2.00 L of a 3.00 M solution? The solute could be anything, and the solvent is usually water. The concentration will always be the conversion factor. M stands for mol/L. mol a. 0 j. 100. b. 1 c. 2 L)(-- d. 3 k. 1.71 1. 0.856 e. 4 m. 10.0 f. 5 -) = L g. 6 h. 234 mol i. 58.44For questions #14-16, you should do your work on a separate piece of paper and then upload it. You should use the GUESS Method of solving for each problem. Also, you have a choice. You can complete both #14 & #15 which are single step question or do the extension problem which is a 2 step problem instead of 14 & 15. You must complete #16. Equations Q = mcAT Q = ±mLf Q = ±mLv Values for Water Clice) = 2.06 J/g°C C(water) = 4.186 J/g°C C(steam) = 1.87 J/g°C Lf (water) = 333 J/g Lv (water) = 2,260 J/g %3D G- S- U- E- S- #14. Calculate the energy required to heat 12.5 g of water from a temperature of 42.2°C to 52.3°C. #15. How much thermal energy is needed in order to turn 15.7 grams of water into ice? Extension: We start with 23.2 grams of ice at 0°C and add heat to the ice. The ice completely melts and then is heated until it reaches 45.3°C. How much energy is added in total? #16. 12.0 g of a metal absorbs 317 J when it is heated from 38.2°C to 82.7°C. What is the specific heat of the…Use the exothermic and endothermic interactive to determine the final temperature of each solution described in the table. Water amount Initial temperature Final temperature 100 mL 20 °C 200 mL 20 °C 100 mL 20 °C Solute amount 5 g 5 g 10 g final temperature: 28.64 ? ? Add 5 g of calcium chloride to 100 mL of water at an initial temperature of 20 °C. final temperature: 24.31 ? Add 5 g of calcium chloride to 200 mL of water at an initial temperature of 20 °C. O O °℃ °℃
- Table 1. Standardization of HCI Solutions (40 pts) Trial Volume of HCI (To the nearest 0.01 mL) Molarity of NaOH (mol/L) Volume of NaOH (To the nearest 0.05 mL) Molarity of HCI* (mol/L) 1 20.00 mL 0.01011 10.40 0.0525 2 20.00 L 0.01011 10.50 0.0530 3 20.00 ml 0.01011 10.50 0.0535 4 20.00 mL 0.01011 10.60 0.0535 5 (If necessary) 20.00 mL 0.01011 10.50 0.0530 *Show calculations. Attach additional sheets of papers. According to experiment, the average molarity of our HCI solution is 0.0531 ± The uncertainty should cover the range of the experimental values. mol/L.< STARTING AMOUNT X A brine solution is 3.50% NaCl by mass. Given that its density is 1.071 g/mL , determine the quantity of liters of solution that contains 5.00 moles of NaCl 5.00 ADD FACTOR 0.1 *( ) 3.50 10 35.45 Question 29 of 33 1.071 0.01 8.03 x 1022 1000 100 ANSWER 7.80 22.99 0.0156 L solution g NaCl/mL g solution mL solution kg solution g NaCl/g solution 58.44 6.022 x 1023 0.0780 moles of NaCl mol NaCl RESET 0.001 2 1 g NaCl 1.56$ F4 An aqueous solution of 2.87 M hydrochloric acid, HCl, has a density of 1.05 g/mL. The percent by mass of HCl in the solution is %. Submit Answer % 5 T G F5 Retry Entire Group 9 more group attempts remaining 6 Cengage Learning | Cengage Technical Support Y H [Review Topics] [References] Use the References to access important values if needed for this question. MacBook Air F6 & 7 gilinent-take F7 8 00 1 DII FB K ( 9 F9 0 L ) O A F10 P Chapter Previous + 11 X Next ☐ Save and Exit OWLY S deles