| 6.99V | 0.14A | 7. | 8.00V | 0.16A | 8. | 9.03V | 0.18A | 9. | 9.98V | 0.20A | 10. | 18.01V | 0.36A | 50-ohm Resistor | V drop | I through | 1.
1,054,848 c. 1,405,888 d. 1,045,828 5. Evaluate: 12xy, when x = 8 and y = 11. a. 31 b. 228 c. 1056 d. 188 6. Write
2 NO + 1O2 ( 2 NO2 4. ____ Al(NO3)3 + ____ Ba(OH)2 ( ____ Al(OH)3 + ____ Ba(NO3)2 5. 1 H2 + 1Br2 ( 2 HBr 6. ____ HCl + ____MnO2 ( ____MnCl2 + ____ H2O + ____Cl2 7. ____ F2 + ____H2O ( ____ O2 + ____HF 8.
"_ .;...........'.1......,.. __..... 11 ........ •_ _........... _ti(i$l,..,;.S! !l'!lIgical, and s ooial a nalys!s. F urthermore, a s ttJ!y Q f n arrative t echnique, i n\:ludinq 1 :he u se o f n arrative v oiee' b rings t ogether· t he , t.hree p reviously e onsidered a spects.
: Se, calculated molar mass 78.93 g] 4) A 0.3528 gram mixture of H2SO4 (molar mass 98 grams) and H3PO4 (molar mass 98 grams) was mixed with a little water and titrated with 9.70 mL of 1.000 M NaOH: 3NaOH(aq) + H3PO4(aq) →3H2O(l) + Na3PO4(aq) 2NaOH(aq) + H2SO4(aq) →2H2O(l) + Na2SO4(aq) What was the percent by mass H2SO4 in the original mixture? [Ans. : 30.6] 5) A 0.9030 gram sample of Z(OH)2 and 20.00 mL of 2.000 M HCl were put into a 100.0 mL volumetric flask and mixed with enough extra water to make 100.0 mL of solution. A 10.00 mL aliquot of this solution was taken and titrated with 17.64 mL of 0.05121 M NaOH. What is the identity of the element Z?
A) +12 B) +4 C) +6 D) -2 E) 0 34. What is the oxidation number of Cl in
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%. 3) Percent recovery for isolation of hydroquinone dimethyl ether % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.16/1.01) x100% = 15.84% That concludes that the percent recovery is 15.84%. Table 2: : Experimental IR peaks compared to literature IR peaks for Benzoic acid Functional groups | Experimental peak (cm-1) | Literature peak (cm-1) | O-H | 3407-2563 | 3400-2564 | C=O | 1689 | 1689 | C-H |
That is why once you add the benzoic acid aqueous solution and the CH2Cl2 in the separatory funnel the benzoic acid moves from the aqueous layer into the methylene chloride organic layer. This will occur until an equilibrium is establish between the aqueous layer and the organic layer; that is there is an equal amount of benzoic acid in the aqueous layer and the organic
Amine Directed Hydroboration: Methodology and Synthetic Application Rebecca A. Murphy Department of Chemistry, University of California, Berkeley, California 94720 United States rmurphy@berkeley.edu ABSTRACT Directed hydroboration has been a topic of interest to practitioners of chemical synthesis for over fifty years. With recent advances in amine directed hydroboration, the field has gained momentum by being featured a series of methodological developments and natural product syntheses. In this paper, we highlight the advances made over this time period, as well as discuss recent applications to natural product synthesis. The discovery of hydroboration by H. C. Brown in 1959 sparked frenzied interest in the field of organoboron chemistry.1 Since this initial discovery, organoboron chemistry has become a mainstay in organic synthesis, and continues to be an important tool. A testament to this are the multiple Nobel prizes awarded for research in this area.