454 Words2 Pages

Warm up questions:
Exercise1： mass of ZnI2 is 2.56g
The molar mass of ZnI2 is 319.18 g/mol
The mole of ZnI2: 2.56g/319.18g/mol=0.00802mol
0.00802mol/(500*10-3L)=0.01604M
0.01604M ZnI2 should appear on the label of the flask.
Exercise 2:
Student 1: 0.43g zinc iodide
The mole of ZnI2: 0.43g/319.18 g/mol= 0.00135mol
0.00135mol/0.01604M=0.084L
0.084L*(1000mL/1L)=84mL
Student 2: 5.0*10-4 moles of zinc iodide.
5.0*10-4mole/0.01604M=0.031L
0.031L*(1000mL/1L)=31mL
Molarity as a Concentration Unit
Exercise 3: a. 2.56g ZnI2/500 mL of solution
2.56g/319.18 g/mol=0.00802mol
0.00802mol/(500*10-3L)=0.016M
b. 0.00512g ZnI2/mL of solution
0.00512g/319.18 g/mol=1.6*10-5 mol
1.6*10-5 mol/(1*10-3L)=0.016M c. 0.00806 moles of ZnI2/500 mL of solution
0.00806mol/(500*10-3)L=0.016M
d. 0.0161 moles of ZnI2/L of solution
0.0161mol/1L=0.016M
Exercise 4: a. The moles of ZnI2: 0.25M*(250*10-3)L=0.0625mol b. 0.25M*(250*10-3)L=0.0625mol
The mass of ZnI2: 0.0625mol*319.18 g/mol=19.95g c. 0.25M*(500*10-3)L=0.125mol
0.125mol*319.18 g/mol=39.9g ZnI2 d. 0.0125mol/0.25M=0.05L
Exercise 5: a. 0.125M*(100*10-3)L=0.0125mol b. 0.0625mol/0.125M=0.5L=500mL
Calculation for preparing the EDTA solution
Exercise 6 a. 1L*0.02M=0.02mol
0.02mol*372.24g/mol=7.4g EDTA b. Exact molarity of 7.4448g /1.00L would be .0200 M Exercise 7 a. 0.5M*100*10-3L=0.05mol acetic acid b. 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.

Related

## Howdy Essay

870 Words | 4 PagesWhat volume of concentrated (18.0 M) sulfuric acid would be required to make each of the following? a. 1.25 L of 6.00 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (6.00 M) x (1.25 L) Vconc=0.417 L b. 575 mL of 0.100 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (0.100 M) x (0.575 L) Vconc=0.00319 L Acid-Base Titrations 5. Calculate the molarity of an HCl solution if 20.0 mL of it requires 33.2 mL of 0.150 M NaOH for neutralization.

## 1.07 Accuracy and Precision

671 Words | 3 PagesData * Below is the table that you will complete for the virtual lab. Either type your results into this table or print the table from the virtual lab (it must be submitted to receive full credit for this assignment. Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 25.31g | 25.40g | 26.03g | Volume of liquid (milliliters) | 8.12mL | 8.26mL | 8.52mL | Mass of graduated cylinder and liquid (grams) | 35.46g | 36.01g | 36.41g | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 42.313g | 40.65g | 40.95g | Volume of water (milliliters) | 48.95mL | 50.03mL | 50.05mL | Volume of water and solid (milliliters) | 53.91mL | 55.04mL | 550.04mL | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 25.95g | 27.62g | 25.67g | Length of solid (centimeters) | 5.250cm | 5.0cm | 4.50cm | Width of solid (centimeters) | 3.0cm | 4.0cm | 3.50cm | Height of solid (centimeters) | 2.50cm | 3.0cm | 2.0cm | Calculations Show all of your work for each of the following calculations and be careful to follow significant figure rules in each calculation. Part I: Density of Unknown Liquid 1. Calculate the mass of the liquid for each trial.

## Chemestry- Finding Pka

956 Words | 4 PagesResults | | | | Trial #1 | Trial #2 | Volume of NaOH at equivalence point | 9.5 mL | 13.7 mL | Volume of NaOH at one-half the equivalence point | 4.75 mL | 6.9 mL | pH at half equivalence point | 5.42 | 5.3 | pKa of unknown acid | 5.42 | 5.3 | Average pKa | 5.36 | 5.36 | Average Ka | 4.4 * 10-6 | 4.4 * 10-6 | Moles of unknown acid | 1.30 * 10-3 moles | 1.87 * 10-3 moles | M of unknown acid | 0.0650 M | 0.0805 M | Average M of unknown acid | 0.07275 M | 0.07275 M | 2. The data table containing all your raw and manipulated data (like the one found on page 47 of your lab manual). Be sure to include your lab partner’s name and unknown number. Unknown acid buret | | | | Trial #1 | Trial #2 | Initial buret reading | 20.5 mL | 21.8 mL | Final buret reading | 40.5 mL | 45.1 mL | Volume of acid added | 20.0

## Lab Report

1509 Words | 7 PagesSiddharth Rajendran Chemistry HL Urea Dissolution Lab Raw Data:- (Expected Values) Change in Enthalpy: 14 kJ mol-1 Change in Gibbs free energy: 6.86 kJ mol-1 Change in Entropy: 69.5 J mol-1 Molar Mass of Urea: 60.06 g mol-1 Heat Capacity: 4.184 J g-1º Data Table 1: To calculate the Enthalpy change Mass of Urea Tablet (g) (±0.01g) | Volume of Water(mL) (±0.05mL) | Initial Temperature (Cº) (±0.2 Cº) | Final Temperature(Cº) (±0.2 Cº) | 3.04 | 50.0 | 21.3 | 17.4 | Initial Observations:- * There was a decrease in temperature at a fast rate. * The temperature of the solution was slowing down continuously but the rate started decreasing. * The Urea dissolved and the rate was decreasing continuously. * The temperature gradually started to increase after almost the Urea present had dissolved. Data Table 2: Mass, Volume and Temperature during Dissolution of Urea (To calculate Keq) Mass of Urea(g) (±0.01g) | Initial Temperature(Cº) (±0.2 Cº) | Final TemperatureCº) (±0.2 Cº) | Initial Volume(mL)(±0.05 mL) | Final Volume(mL)(±0.05 mL) | 3.76 | 21.4 | 22.9 | 5.02 | 7.14 | Processing Raw Data * Determining the Final temperature of dissolution of Urea in the Styrofoam cup.

## Ionic Reactions Essay

882 Words | 4 PagesExperiment 8 Ionic Reactions Karen Curry 11/4/2013 10:30am Chemistry 12011K Abstract The purpose of this experiment is to determine the precipitate when two aqueous solutions are mixed together. Then, writing an ionic equation to determine the identity of the soluble or insoluble compound. Experiment and observation I located the equipment I needed from the LabPaq and set up the 96 well plate. I set the chemical pipetes in the 26 well plate in the order that I was going to use them. Going horizontally, I started with cobalt (II) nitrate and placed 2 drops into seven of the A row wells of the 96 well plate.

## Unit 8 Chemistry Kaplan

782 Words | 4 PagesTo find out how much acid was neutralized by single doses of each antacid we have to subtract the answers from question one from the moles of HCl (.004mol). Maalox: .004 -.001205= .002795 mol Tums: .004 - .00112 = .002880 mol Mylanta: .004 - .001 = .003 mol CVS brand: .004 - .000995 = .003005 mol Rennies: .004 - .00122 = .002780 mol Divide the number of moles of acid neutralized in each case by that antacid's mass. This gives you the moles of acid neutralized per gram of antacid, which will allow you to judge the strongest and weakest antacids on a per weight basis. Maalox: 24.1ml/20.0g = 1.21ml/g .002795 mol base / 20.0g = .00013975 mol base/pill Tums: 22.4ml/21.0g = 1.07ml/g .002880 mol base / 21.0g = .0001371 mol base/pill Mylanta: 20.0ml/18.0g = 1.11ml/g .003 mol base / 18.0g = .0001666 mol base/pill CVS brand: 19.9ml/18.3g = 1.09ml/g (.003005 mol base)/ (18.3g pill) = .0001642 mol base/pill Rennies: 24.4ml/ 17.5g = 1.39ml/g (.002780 mol base)/ (17.5g pill) = .0001588 mol base/pill

## Limiting Reactants Lab Report

864 Words | 4 PagesTest Tube 4 1.5mL/1 x 1L/1000mL x .5/1L = 7.5 x10-4 Test Tube 5 2.0mL/1 x 1L/1000mL x .5/1L = 1.0 x10-3 Determine LR Test Tube One (.001mol NaI / 1.5 x 10-4mol Pb(NO3)2) = 6.66 (1 / 1) = 1 6.66 > 1 Pb(NO3)2 is the Limiting Reactant Test Tube Two (.001mol NaI / 2.5 x 10-4mol Pb(NO3)2) = 4 (2 / 1) = 2 4 > 2 Pb(NO3)2 is the Limiting Reactant Test Tube Three (.001mol NaI / 5 x 10-4mol Pb(NO3)2) = 2 (2 / 1) = 2 2 = 2 Test Tube number Three is the Exact stoich ratio Test Tube Four (.001mol NaI / 7.5 x 10-4mol Pb(NO3)2) = 2 (2 / 1) =

## Chem 130 Questions

5759 Words | 24 PagesComparing the rate of appearance of C and the rate of disappearance of A, we get[pic]. A) [pic] B) [pic] C) [pic] D) [pic] E) [pic] Answer: A Diff: 1 Page Ref: Sec. 14.2 A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) →B(g). The following data are obtained for [A] as the reaction proceeds: [pic] 5) The average rate of disappearance of A between 10 s and 20 s is __________ mol/s. A) [pic] B) [pic] C) [pic] D) 454 E) [pic] Answer: A Diff: 1 Page Ref: Sec.

## Crime Lab Essay

418 Words | 2 PagesThen prepare the unknown sample by pipetting 1 ml of the unknown solution into the cuvette and mix with 1 ml of dilute ferric nitrate. Then after that you calculate the concentration for each of the solutions. Cuvette # | 25 mg/dl Standard | H2O ml | Dilute Iron III Nitrate (ml) | .039 M HNO3 (ml) | Concentration | 1 | 0 | 1 | 0 | 1 | 0 | 2 | 0.1 | 0.9 | 1 | 0 | 0.09 | 3 | 0.3 | 0.7 | 1 | 0 | 0.287 | 4 | 0.5 | 0.5 | 1 | 0 | 0.407 | 5 | 0.7 | 0.3 | 1 | 0 | 0.706 | 6 | 1 | 0 | 1 | 0 | 1.316 | When then use the calibration curve to determine the concentration of the unknown and we get .747. IV. As you can see from our data as the amount of H2O decreases and the 25mg/dl standard increases as does our concentration for the salicylic acids.

## Lab 6: Heat of Reaction and Hess’s Law

630 Words | 3 PagesData Table(s): Reaction equation Mass of solid NaOH Initial Temp. Final Temp Delta T NaOH 2.0351 g 23.6 C 31.8 C 8.2 C HC2H3O2 1.9966 24.8 C 41.6 C 16.8 C NaOH + HC2H3O2 25.4 C 32.8 C 7.4 C Calculations/Results: 2.0g NaOH/ 40.0g NaOH= .05M NaOH 174.8kj + -351.12kj = -176.32 174.8 + -351.12) - (154.66) / 154.66 = 2.14% Questions/Problems:Pre-Lab 1.) 2Na+(aq) + OH-(aq)+ HC2H3O2(aq) -> H2O(l) + 2NaOH(s)+C2H3O2-(aq) 2. )Na+(aq) + OH-(aq) -> NaOH(s) +NaOH(s) + HC2H3O2(aq) -> H2O(l) +

### Howdy Essay

870 Words | 4 Pages### 1.07 Accuracy and Precision

671 Words | 3 Pages### Chemestry- Finding Pka

956 Words | 4 Pages### Lab Report

1509 Words | 7 Pages### Ionic Reactions Essay

882 Words | 4 Pages### Unit 8 Chemistry Kaplan

782 Words | 4 Pages### Limiting Reactants Lab Report

864 Words | 4 Pages### Chem 130 Questions

5759 Words | 24 Pages### Crime Lab Essay

418 Words | 2 Pages### Lab 6: Heat of Reaction and Hess’s Law

630 Words | 3 Pages