ofcr in 1 10. mL of a solution containing 3.22 g of CaCl2 is Mb. 0.0290 M c. 0.0
ID: 574334 • Letter: O
Question
ofcr in 1 10. mL of a solution containing 3.22 g of CaCl2 is Mb. 0.0290 M c. 0.00319 M d. 0.528 M e. 0.354 M m 11-The S What volume of a 0.550 M solution of potassium hydroxide can be made with 14.3 g of potassium hydroxide? Ignore significant figures for this problem.) a 26.0 mL (5463 12. d. 2158 mL e. 2.16 mL c. 255 mL 39+16t1 3. What volume of I8.0 M sulfuric acid must be used to prepare 15.SL of 0.121 M HSO.? (lgnore figures for this problem.) e. none of these 04 mL b.141 mLc. 1.88L d. 128.1 mL 2.00 M solution of AgNOslaq) called solution B. You mix these solutions together, making solution C Calculate the concentration (in M) of Na* ions in solution 14. You have 3.00 L of a 3.48 M solution of NaCl/a) called solution A. You also have 2.00 L of a .3.48Mc 5.22 M d. 1.10M e. 2.09 M You mix 100.0 mL of a 0.100 M NaOH solution and 150.0 mL of a 0.104 MHCl solution. Determine the concentration of H in the final mixture after the reaction is complete. 15. 0.0624 M b. 0.104 M c. 0.156 M d. 0.204 M e. 0.0224 M Which of the following solutions contains the smallest number ofions? a 16. 539.8 mL of 1.0 M lithium nitrate31xIx 189 6 mL of2.0 Miron III) chloride b x n y 4 e At least two of the above contain the smallest number of ions. 7n If a 74.6-g sample of ammonium nitrate is dissolved in enough water to make 315 mL of a. 0.932 rn: the density of this solution is 0.92g/ml. what will be the molality? c. 0.294 m d. 3.22 m e. none of these 8. The bonds between hydrogen and oxygen in a water molecule can be characterized as a hydrogen bonds. b. London forces. c. intermolecular forces. e. forces.Explanation / Answer
Ans 11
Moles of CaCl2 = mass/molecular weight
= 3.22g / 110.98g/mol
= 0.0290 moles
Volume of solution = 110 mL x 1L/1000 mL
= 0.110 L
Molarity = moles/volume = 0.0290 moles/0.110 L
= 0.264 M
Option A is the correct answer
Ans 15
Moles of NaOH = molarity x volume
= 0.100 moles/L x 0.100 L
= 0.0100 moles
Moles of HCl = 0.104 x 0.150 = 0.0156 moles
balanced chemical equation
NaOH + HCl ----> H2O + NaCl
From the stoichiometry of the reaction
1 mol NaOH reacts with = 1 mol of HCl
0.0100 mol NaOH reacts with = 0.0100 mol of HCl
NaOH is limiting reactant
Excess reactant = HCl
HCl left after the reaction = initial - reacted
= 0.0156 - 0.0100 = 0.0056 moles
Moles of H+ in the final mixture = HCl left after the reaction
Total volume = 100 + 150 = 250 mL = 0.250 L
Molarity = 0.0056 moles/0.250L
= 0.0224 M
Option E is the correct answer
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