the questions are answerd: please solve 19-20-21 step by step and explain why 22
ID: 557576 • Letter: T
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the questions are answerd:
please solve 19-20-21 step by step
and explain why 22-23 answers ?
19. Wha ms n -4.9 × 10- a. 2.1 g b. 4.9 g 0.95 g o-6= 4 e, the vertical portion around the equivalence point is shorter than n 0.00022 g 136.tt.) e. 20. In a strong acid/weak bas a strong acid strong base titration curve, so many indicators would not be suitable for a the equivalence point. Which one of the following (CH))>N, with 0.100 M hydrochloric acid, HCI? indicators would be the best choice for the titration of 100.0 mL of 0.100 M methyl orange, red to yellow, pH 3.1-4.4 chlorophenol red, yellow to red, pH 4.8-6.4 alizarin yellow R, yellow to red, pH 10.1-12.0 indigo carmine, blue to yellow, pH 11.4-13.0 c. d. 21. The solubility of Cu(OH), is 1.72x 10 *g per liter at 20°C. Calculate the solubility product for Curo a. 7.3 x 10-30 b. 5.2 x 10-6 c. 2.0 × 10-20 d1.6 x 10-3 e. 4.0 ×10-4 for Fe(IO,), is 10-". We mix two solutions of equal volume, one containing 10-2 MFe and one containing 10-2 MIO,-ions at 25°C. Which one of the following statements is true? a. No precipitate forms, because Op K A precipitate forms, because d. A precipitate forms, because QpK e. None of the other statements is true. 23. Which of the following solubility product expressions is incorrecet?Explanation / Answer
19)
The salt dissolves as:
CaSO4 <----> Ca2+ + SO42-
s s
Ksp = [Ca2+][SO42-]
4.9*10^-5=(s)*(s)
4.9*10^-5= 1(s)^2
s = 7*10^-3 M
Molar mass of CaSO4 = 1*MM(Ca) + 1*MM(S) + 4*MM(O)
= 1*40.08 + 1*32.07 + 4*16.0
= 136.15 g/mol
Molar mass of CaSO4= 136.15 g/mol
s = 7*10^-3 mol/L
To covert it to g/L, multiply it by molar mass
s = 7*10^-3 mol/L * 136.15 g/mol
s = 0.95 g/L
So, in 1 L, mass dissolved = 0.95 g
Answer: d
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