Chemical Reactions 11) ethanol (400 was diluted to 400 L glvinga concentration o
ID: 490436 • Letter: C
Question
Chemical Reactions 11) ethanol (400 was diluted to 400 L glvinga concentration of oot00M The concentration of the original solution was p. 2 B)4.00 12 A 175 mL sample of an D) 1.60 neutralization. acetic acid (CH3co2h0 solution required 29.6 mL of 0.250 M NaOH for The concentration of acetic acid was E) 0.200 0.21 13) Which of the following is soluble in water at 25 C? A) Fe3(PO4)2 B) Fes. D) 0.42 Fe(NO3)2 D) Fe(OH)2 The point in a titration at which the indicator changes is called the A) volumetric B) indicator point standard point D) equivalence point E) 15 What is the molarity of an aqueous solution containing 753 g of glucose C6H1206 in 35.5 mL of solution? A) 352 B) 1.85 C) 0.197 D) 2.12. E) 11.8 16) What mass (g) of AgBr is formed when 35.5 mLofo.184 MAgNO3 is treated with an excess of aqueous hydrobromic acid? A) 1.23 B) 53.6 D) 188 E) 1.44 17) zinc is more active than cobalt and iron but less active than aluminum. Cobalt is more active than nickel but less active than iron. Which of the following correctly lists the elements in order of increasing activity? A) Co Ni Fe Al B) Zn Al Co Ni Fe c) Ni Fe Co Zn Al D) Fe Ni Co Al Zn E) Ni Co Fe Zn Al 18) Ammonia is a A) strong base B) weak base C) weak acid D) strong acid salt 19) of the following elements, is the most easily oxidized. A) fluorine B) gold C) aluminum D) oxygen E) nitrogen 20 The balanced molecular equation for complete neutralization of H2soy by KoH in aqueous solution is 2H+ (aq) 2KoH (aq) 2H20 2Kt (aq) H2sos (aq) 20H- (aq) 2H20 (I) song 2- (aq) 21) Calculate the number of grams of solute in 500.0 mL of 0.189 M KoH. A) 5.30 x 103 B) 148 1.68 D) 1.68 x 10-3 EO 5.30Explanation / Answer
Before dilution After dilution
M1 = M2 = 0.04M
V1 = 400ml V2 = 4L = 4000ml
M1V1 = M2V2
M1 = M2V2/V1 = 0.04*4000/400 = 0.4M>>>C
12. CH3COOH + NaOH --------------> Ch3COONa + H2O
1 mole 1mole
CH3COOH NaOH
M1 = M2 = 0.25M
V1 = 17.5ml V2 = 29.6ml
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
M1 = M2V2n1/V1n2
= 0.25*29.6*1/17.5*1 = 0.42M >>>>>>D
13. C
all nitrate salts are soluble in water.
14. D
15. molarity = W*1000/G.M.Wt * volume of solution ml
= 75.3*1000/180*35.5 = 11.8 >>>>>>answer>>>>>>>>E
17.A
18.B
19. C
20. A
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