c. 1.6 x 103 M 28) The concentration of sulfate in a sample of wastewater is to
ID: 1035008 • Letter: C
Question
c. 1.6 x 103 M 28) The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of lead mir forming the insoluble lead sulfate (303.3 g/mol) according to the balanced equation given beloww The solid lead sulfate is dried, and its mass is measured to be 0.7256 g. What was the concentration of sulfate in the original wastewater sample? Page 1 5 a. 0.02392 M b. 2.2 M e. 41.79 M d. 4.179 M 22M 29) 2n reacts with hydrochloric acid. Zn(s) + 2 HCl(aq)-. Znckaq) + H2(g) What volume of 3.00 M HC? (aq) will react with 32.69 g Zn(s)? a. 0.333 L b. 0.502 L c, 32.69 L d. 2.33 L e. 3.00 L 30) A mass of 0.4113 g of an unknown acid, HA, is titrated with NaOH. If the acid reacts with 28.10 mL of 0.1055 M NaOH(aq), what is the molar mass of the acid? a. 2.965 x 10g/mol b. 9.128 g/mol c. 337.3 gnol d. 138.7 g/mol e. 820.7 g/mol 31) The reaction of HCI with NaOH is represented by the equation HCl(ag)+ NaOH(aq)NaC(aq)+ H:O0) What volume of 0.825 M HCl is required to titrate 25.0 mL of 0.525 M NaOH? c.21.9 mL d. 25.0 mL e. 52.5 mL a. 18.1 mL b. 15.9 mL I. Absorbance is directly proportional to the intensity of the incident light. 2. Absorbance is directly proportional to the analyte concentration. 3. Absorbance is directly proportional to the path length of the light. a. I only b. 2 only c. 3 only d. 2 and 3 e. 1,2, and 3 32) Which of the following statements is/are CORRECT?Explanation / Answer
Answer for 28 is "a"= 0.023M
303.3 gm of PbSO4 precipitates 96 g of SO4 2-
Hence 0.7256 g * 96 / 303.3 gm
= 0.023 M
Answer for 29 is "a" = 0.333M
32.69 g * 1 mole of Zn / 65.41 g = 0.5 mole of zinc
In order to react 0.5 M of zinc we need
0.5 mole Zn * 2M HCl/ 1 M Zn= 1M HCl
1M HCl * 1L solution/ 3M HCl = 0.333 M
Answer for 30 is "d" = 138.7 g/ mol
No.of mole of NaOH = concentration* volume
= 0.1055 * 928.1/1000
= 2.96 * 10^-3
No.of moles = mass / molar mass
Molar mass = mass / no.of moles
= 0.4113/2.96*10^-3
= 138.7 g/ mol
Answer for 31 is "b" = 15.9 mL
HCL =NaOH
M1 *V1 = M2 * V2
V1 = 0.525*25/0.825=15.9 mL
Answer for 32 is "d" = 2&3
Axcording to Beers law and Lambert's law.
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