Metal Sample Large 32.30 Triple-beam Balance Small Mass Beaker (g) Mass of Metal + Beaker (g) 110.74 Initial Volume (mL) Final Volume (mL) 32.30 67.22 63.8 55.5 70.5 58.7 Metal Sample Large 32.56 Electronic Balance Small Mass Beaker (g) Mass of Metal + Beaker (g) 89.06 Initial Volume (mL) Final Volume (mL) 32.56 53.71 59.9 55.2 65.0 57.1 Metal Sample Small 32.5760 32.5760 Mass of Metal + Beaker (g) 56.4270 56.5776 Analytical Balance Large Mass Beaker (g) Initial Volume (mL) Final Volume (mL) 56.2 56.2 58.2 57.8 The following data was obtained in the lab. Calculate the average density of the metal obtained on the electronic balance. 10.0 11 11.01 11.1

Metal Sample Large 32.30 Triple-beam Balance Small Mass Beaker (g) Mass of Metal + Beaker (g) 110.74 Initial Volume (mL) Final Volume (mL) 32.30 67.22 63.8 55.5 70.5 58.7 Metal Sample Large 32.56 Electronic Balance Small Mass Beaker (g) Mass of Metal + Beaker (g) 89.06 Initial Volume (mL) Final Volume (mL) 32.56 53.71 59.9 55.2 65.0 57.1 Metal Sample Small 32.5760 32.5760 Mass of Metal + Beaker (g) 56.4270 56.5776 Analytical Balance Large Mass Beaker (g) Initial Volume (mL) Final Volume (mL) 56.2 56.2 58.2 57.8 The following data was obtained in the lab. Calculate the average density of the metal obtained on the electronic balance. 10.0 11 11.01 11.1

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter1: Basic Concepts Of Chemistry

Section: Chapter Questions

Problem 2RCYU: A student checked the accuracy of two standard top-loading balances by testing them with a standard...

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A student checked the accuracy of two standard top-loading balances by testing them with a standard 5.000-g mass. The results were as follows: Balance 1: 4.99 g, 5.04 g, 5.03 g, 5.01 g Balance 2: 4.97 g, 4.99 g, 4.95 g, 4.96 g Calculate the average values for balances 1 and 2 and calculate the percent error for each. Which balance is more accurate?
1. 500.0 mL 0.2 M KI2. 500.0 mL 0.2 M KCl3. 500.0 mL 0.1 M K2S2O84. 500.0 mL 0.1 M K2SO45. 500.0 mL 4.0 mM Na2S2O3 (from Na2S2O3∙5H2O)pls compelete the 2nd table given the data
Show all the work and setps. Perform the following calculations using your calculator and then express the answer in the correct units and number of significant figures. 7.945 J + 82.3 J − 0.02 J 0.0012 m − 0.000 45 m − 0.000 11 m 500 g + 432 g + 2 g 31.2 kPa + 0.0035 kPa − 0.147 kPa 312 dL − 31.2 dL − 3.12 dL 1701 kg + 50 kg + 43 kg
Why should the selected rule(s) be followed? A residue left on the balance will contaminate the balance pan surface and anything that comes into contact with it afterwards. There may be discrepancies in accuracy between individual balances. Objects at any temperature other than room temperature can create air currents which can affect the balance pan, causing the measured mass to be higher or lower than the true mass. Weighing wet items will cause the displayed mass to be greater than the actual mass of the object.
Chemistry Fraction no. Elution Vol. (mL) Abs (λ = 410) Fraction no. Elution Vol. (mL) Abs (λ = 410) 1 (PB pH 6.0) 1 0.0050 11 11 0.0041 2 2 0.0032 12 12 0.0060 3 3 0.0040 13 13 0.0064 4 4 0.0064 14 14 0.0063 5 5 0.0067 15 15 0.1109 6 6 0.0067 16 16 2.8902 7 7 0.0050 17 17 0.8500 8 (PB pH 8.0) 8 0.0049 18 18 0.2019 9 9 0.0052 19 19 0.0378 10 10 0.0050 20 20 0.0073 Use the data above to plot an elution profile, in which the absorbance is on Y-axis, and elution volume (volume = fraction number) is on X-axis. The elution profile must indicate at (1) which point the pH change occurred, (2) the fraction(s) which contain the highest myoglobin content, and (3) the maximum absorbance detected.
In the UV assay of paracetamol you are required to weigh tablet powder “…containing 0.150 g of Paracetamol…”. If the stated content of paracetamol per tablet is 500 mg and the weight of 20 tablets is 11.223 g then the amount of paracetamol tablet powder to be weighed is
Sample Physical Appearance 0.1 M Na2CO3 0.1 M CaCl2 Na2CO3 + CaCl2 mixture Part II filtrate + CaCl2
8.3m x 4.0m x 0.9823m = 32.61236 m^3 Explain why and how you would make the answer 2 sig. Figs.
A group of students compiled the data shown in data table 1 below. What is the exact calculated order for H+(aq) to 2 decimal places? Exp. # [IO3-]0 (M) [I-]0 (M) [H+]0 (M) Time (s) 1 0.005 0.05 2 x 10-5 22.12 2 0.010 0.05 2 x 10-5 86.84 3 0.005 0.10 2 x 10-5 5.35 4 0.005 0.05 4 x 10-5 2.65 Question 6 options: 2.82 -2.51 3.34 2.71 -2.58 3.12 2.91 3.49 3.41 3.23 3.06 2.64
Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
1. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(8.50mL-8.74mL)2+(9.72mL-8.74mL)2+(8.00mL-8.74mL)2/(3-1)]½ 2. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(10.69-11.52)2+(11.80-11.52)2+(12.07-11.52)2/(3-1)]½
pls complete the 2nd table given the data1. 500.0 mL 0.2 M KI2. 500.0 mL 0.2 M KCl3. 500.0 mL 0.1 M K2S2O84. 500.0 mL 0.1 M K2SO45. 500.0 mL 4.0 mM Na2S2O3 (from Na2S2O3∙5H2O)