Solubility Curves of Selected Compounds 150 140 130 120 110 100 NaNO 90 80 70 60 50 NHC KCI 40 Naci 30 20 RCIO 10 O 10 20 30 40 50 60 70 80 90 100 Temperature (C) The solubility of ammonium chloride at 70 °C is O 34 g/100 g water O 39 g /100 g water O 47 g/ 100 g water O 60 g/100 g water 4 5 6. 7. 2. FONDN NH) Grams of solute

Solubility Curves of Selected Compounds 150 140 130 120 110 100 NaNO 90 80 70 60 50 NHC KCI 40 Naci 30 20 RCIO 10 O 10 20 30 40 50 60 70 80 90 100 Temperature (C) The solubility of ammonium chloride at 70 °C is O 34 g/100 g water O 39 g /100 g water O 47 g/ 100 g water O 60 g/100 g water 4 5 6. 7. 2. FONDN NH) Grams of solute

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

100.0 mL of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)2 MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 -): (5.831 g Mg(OH)₂ (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration a. acid i. 0.2000 mol Mg(OH)2 (-- mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI p. 0.01000 b. base c. HCI NaOH j. 1.000 k. 0.005000 q. 0.001000 L HCI r. 58.31 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000
pls answern 9 and 10
You have an ointment that contains 40% (w/w) of ingredient Y and another ointment which contains 5% (w/w) of ingredient Y. You wish to make 300g of ointment which contains 25% (w/w) of ingredient Y. Which ONE of the following corresponds to the correct amounts of 40% (w/w) and 5% (w/w) ointments that you will need to mix? A. 200.1 g of the 40% ointment; 99.9g of the 5% ointment B. 192.7 g of the 40% ointment; 107.3g of the 5% ointment C. 171.4 g of the 40% ointment; 128.5g of the 5% ointment D. 138.8 g of the 40% ointment; 161.2g of the 5% ointment E. 97.5 g of the 40% ointment; 202.5g of the 5% ointment
of F. Hydrazine, N,H, reacts with oxygen to form nitrogen gas and water. If 3.75 g of N,H, reacts with excess oxygen and produces 0.750 L of N, , at 295 K and 1.00 atm, what is the percent yield of (DOʻHT + (3)°N – (3)°0 + (be)*H’N the reaction? percent yield: MacBook Pro & 4. 9. 7. 6. P. H. K. B M. N
15. (5 points) Calculate the vapor pressure (in torr) at 25 °C of a solution containing 49.8 mL ethylene glycol (HOCH₂CH₂OH; a non-volatile compound; density=1.11g/mL) and 285.2 g water. The vapor pressure of pure water at 25 °C is 23.8 torr. Imol 99.8 ML 55.2789 102,079 1.11g ML =
Fill in the blank
(10) What is the boiling point of a solution of 12.0 g of urea (CH4N2O) in 165.0 g of water? (Ko (H2O) = 0.52 C/m). (11) A 5.17 molal aqueous nitric acid solution was prepared. (Density of nitric acid = 1.51 g/mL, density of water = 1.0 g/mL, and the MW of nitric acid = 63 g/mol)What is the concentration (%"/v) of the solution? (12) What is the osmotic pressure of a solution that contains 13.7 g of propyl alcohol (C3H;OH) dissolved in enough water to make 500. ml of solution at 27C?
12:47 Question 1 of 10 Submit How many moles of Na,SO, are in a 0.140 aqueous solution that was prepared with 2.30 kg of water ? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 1 45.7 kg Na,SO4 322 g Na,SO, 6.022 x 1023 g water 2.30 kg water 0.140 mol Na,SO, 142.04 0.609 0.322 609 Tap here or pull up for additional resources
Solve all parts otherwise I will downvote
9) How many grams of KNO3 are required to make 100.0 mL of a 0.531 M KNO3 solution? (Answer: 5.37 g KNO3) Volume of soIution (V) = 100.0OML= 0.100L Unifs of mojanty= mol /L %3D Molanty of soluhon(M)=0.531H= 0.531 mol/L Nu -W n= moles of soIule = mass/molar mass Molar mass of KNO3= 101.1 glmol Mass of KN03= V = Volume oF SOluion in litres H= (mass/ molar mass) (1/v) Mass = M molar mass V Mass-o. 531 mol/L 101. 1g/mol 0. 100 L M= (mass/molar mass) × (1/V) Mass = 5.37D. 10) How many milliliters of the solution prepared in the question above would be required to make 250.0 mL of a 0.1000 M KNO3 solution? (Answer: 47.1 mL)
What is the boiling point in ºC of a solution made with 25 g of C6H12O6 (a nonelectrolyte) in 100 g of water? (Kb for H2O is .51) Group of answer choices .78 106.8 100.7 101.4
Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality. • Molality is the number of moles of solute dissolved in a kilogram of solvent. molality = moles of solute or molal, m kg of solvent Sample Problem #1: What is the molality of the example sucrose solution? First Step: Known and Unknown Quantities known unknown mass of sucrose = 0.030464 g C12H2201 molality C12H2201 = ? m mass of water = 0.981256 g H,0 Second Step: Formula molality of C12H22O11 = mol of C12H22O1 1 kg of H,0 Third Step: Rearranging, intermediate calculations, and unit conversions No rearranging required. Intermediate calculations: molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g # mol C12H2201 = 0.030464 g C12H,O1 x 1 mol C,,H„01 = 0.000089 mol C12H22O1 1 342.299 g C12H2201 Unit conversions: 0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,0 1 1000.00 g Fourth Step: DELL 「田