2,4-dinitrophenolDraw the reaction equation showing the specific product synthesized.H.OHNO21) Et3N2)NO22,4-dinitrophenolNO22. 9.48.сл0NO20NO2bis (2,4 dinitrophenyl) oxalateShow the calculations determining theoretical and percent yields (the phenol is limiting reactant).mm =184.1064.784gbunbood bas moo

The given molecule 2,4 - dinitrophenol can be synthesised from phenol by nitration reaction, which…

Answered: 2,4-dinitrophenol Draw the reaction…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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GRAVIMETRIC ANALYSIS OF CALCIUM CARBONATE-DATA COLLECTION AND ANALYSIS 8. Fill in the table below; answer YES or NO for each error. Could it affect your Would it affect your calculation of theoretical yield? Possible error measurement of actual yield? "Wet" CaCl2 2H2O Had "difficulty" with the balance Didn't use/read the pump correctly when measuring Na2CO3(aq) (added more or less than you thought) Molarity on the Na2CO3(aq) is incorrect Let the mixture boil on the hotplate Had fine (small) crystals rather than large ones after digesting the precipitate Had a hole in your filter paper, resulting in a cloudy liquid when filtering Didn't heat the solid in the oven long enough
What mass of FeS is required to produce 87g Fe2O3 with excess O2 by this reaction 4FeS + 7 O2 —> 2Fe2O3 + 4SO2
H3C, HO H3C° CH3 Reactant A Reactant B Product 108.14 g/mol 102.09 g/mol 150.18 g/mol 0.108 g 0.112 g 0.120 g
If you are given 1.35g Fe(NO3)3 and 0.00597 mol of Cl2 to conduct the gas forming reaction below 2Fe(NO3)3+2Cl2=2FeCl2+6NO2+3O2. What it's the limiting reagent? What is the theoretical yield of O2 formed in this reaction in moles. When the reaction was run only 0.00256 mol O2 are collected. What is the % yield for this process
4. Consider the reaction MnO2 + 4HCI→ MnCl2 + Cl2 + H2O If 86.3 L of MnO2 and 48.2L of HCl react, a) Which reagent will be used up first? Which is the limiting reagent? b) How many grams of Cla will be produced?
Please see pics. This table looks confusing please help.
How do I find the amount of mols dispensed in NaOH, with the correct amount of significant figures?
#25y of a 23% by mass magnesium chloride (MW= 95.21 g/mole) solution is reacted with 125.0 mL da125M solution of silver nitrate (MW= 169.87g/mole) determine the mass of silver chloride (MW = 143.32 p/mole) produced
2. For the following reaction, Na sodium phenoxide C6H5ONa 116.09 g/mol butyl iodide C4Hgl 184.02 g/mol d = 1.617 g/mL butyl phenyl ether C10H140 150.22 g/mol + Nal a. How many equivalents of butyl iodide are used when 2.0 g of sodium phenoxide is combined with 7.8 mL of butyl iodide? b. How many grams of sodium phenoxide are required if 5.0 mL of butyl iodide is used? Be sure to use the equivalents from part a.
30) What is the percent yield for the reaction below? Only 28.16 g of CO2 are observed to have formed even though stoichiometry would predict the production of 225.3 g of CO2 since 4.000 X100 moles of C3H18 and 8.000 moles of O2 were used. 2 C8H18+ 25 O2 → 16 CO2 + 18 H2O A) 12.50% B) 25.00% C) 50.00% 20.00%
5. When 5.0 moles of potassium metal reacts with an excess of H₂0, 3.5 g of H2 are produced. 2 K (s) + 2 H20 (1) 2 KOH (aq) H₂ (g) a. Calculate the number of moles of water needed to react with all of the potassium metal. (Ans: 5 mol) + b. Calculate the theoretical yield of H2, in grams, formed when all 5.0 moles of potassium metal are consumed. (Ans: 5.04 g) c. Calculate the percent yield of the reaction, given 3.5 g (actual yield) are produced. (Ans: 69.4%)
What quantity of carbon dioxide, CO2 (in grams) would be required to convert 5.60 g of 2-bromopropane to the carboxylic acid named isobutyric acid? [Enter a value with 3 sig figs and do not include any units] 1. Mg, Et,O 2. CO2 3. Н" он Br 2-bromopropane (MW = 122.9 g/mol) isobutyric acid (MW = 88.05 g/mol)

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