Ge As (1) The element with an electron configuration of 1s²2s²2p®3s²3p64s²3d5 is in group and period (2) The element with an electron configuration of 1s22s22p5 is in group and period Submit Answer Retry Entire Group 9 more group attempts remaining (Previous Next Email Instructor Save and Exit

Ge As (1) The element with an electron configuration of 1s²2s²2p®3s²3p64s²3d5 is in group and period (2) The element with an electron configuration of 1s22s22p5 is in group and period Submit Answer Retry Entire Group 9 more group attempts remaining (Previous Next Email Instructor Save and Exit

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter2: Atoms, Molecules, And Ions

Section: Chapter Questions

Problem 85QAP

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A materials engineer has filed for a patent for a new alloy to be used in golf club heads. The composition by mass ranges from 25 to 31% manganese, 6.3 to 7.8% aluminum, 0.65 to 0.85% carbon, and 5.5 to 9.0% chromium, with the remainder being iron. What are the maximum and minimum percentages of iron possible in this alloy? Use Figure 2.12 to snake a prediction about how the density of this alloy would compare with that of iron; justify your prediction.
Define and e xplain the differences between the following terms. a. law and theory b. theory and experiment c. qualitative and quantitative d. hypothesis and theory
You take three compounds, each consisting of two elements (X, Y. and/or Z), and decompose them to their respective elements. To determine the relative masses of X, Y. and Z. you collect and weigh the elements, obtaining the following data: Elements in Compound Masses of Elements 1. X and Y X = 0.4 g, Y = 4.2 g 2. Y and Z Y = 1.4 g, Z = 1.0 g 3. X and Y X = 2.0 g, Y = 7.0 g a. What are the assumptions needed to solve this problem? b. What are the relative masses of X, Y. and Z? c. What are the chemical formulas of the three compounds? d. If you decompose 21 g of compound XY, how much of each element is present?
The early alchemists used to do an experiment in which water was boiled for several days in a sealed glass container. Eventually. some solid residue would appear in die bottom of the flask, which was interpreted to mean that some of the water in the flask had been converted into earth. When Lavoisier repeated this experiment, he found that the water weighed the same before and after heating, and the mass of die flask plus the solid residue equaled the original mass of the flask. Were the alchemists correct? Explain what really happened. (This experiment is described in the article by A. F. Scott in Scientific American, January 1984.)
5. A circulated U.S. quarter has a mass of 5.59 g. Express this mass in milligrams. 5.59 × 103 mg 5.59 mg 55.9 mg 559 mg
Several samples of methane gas, the primary component of natural gas, are decomposed into carbon and hydrogen. The masses of the carbon and hydrogen are then weighed, and the results are tabulated as shown here. Which of these does not follow the law of constant composition? a. 4.0 grams hydrogen and 12.0 grams carbon b. 1.5 grams hydrogen and 4.5 grams carbon c. 7.0 grams hydrogen and 17.0 grams carbon d. 10 grams hydrogen and 30 grams carbon
Am I right? Calculate the mass percentage of water in the hydrate. Show the calculations. Sample 1 : 60.87% Mass of hydrate : 0.46g Mas of water : 0.28g Mass percentage of water in the hydrate = 0.28g/0.46g x 100 = 60.87% Sample 1 2 3 Calculations for sample 1 Mass of aluminum dish 1.24g 1.26g 1.24g - Mass of aluminum dish + Epsom salt 1.70g 1.94g 1.80g - Epsom salt, original (hydrated) mass 0.46g 0.68g 0.56g 1.70g – 1.24g = 0.46g Mass of aluminum dish + Epsom salt after heating 1.42g 1.62g 1.61g Epsom salt, heated (dehydrated) mass 0.18g 0.36g 0.37g 1.42g- 1.24g = 0.18g Mass of water 0.28g 0.32g 0.19g 0.46g – 0.18g = 0.28g Moles of magnesium sulfate 0.0015mol 0.0030mol 0.0031mol 0.18g x (1mol/120.37g)= 0.0015mol Moles of water 0.016mol 0.018mol 0.011mol 0.28g x (1mol/18.02g) = 0.016mol Ratio of moles water: moles of magnesium sulfate 32:3 18:3 11:3.1…
a.) I=R, II = Rb.) I = S, I = Rc.) I=R, II = Sd.) I= S, I = S
F2(g)+2e-→2F-(aq) 2.866 O3(g)+2H+(aq)+2e-→O2(g)+H2O(l) 2.076 Co3+(aq)+e-→Co2+(aq) 1.92 H2O2(aq)+2H+(aq)+2e-→2H2O(l) 1.776 Ce4+(aq)+e-→Ce3+(aq) 1.72 Au+(aq)+e-→Au(s) 1.692 PbO2(s)+4H+(aq)+SO42-(aq)+2e-→PbSO4(s)+2H2O(l) 1.691 MnO4-(aq)+8H+(aq)+5e-→Mn2+(aq)+4H2O(l) 1.507 Au3+(aq)+3e-→Au(s) 1.498 Cr2O72-(aq)+14H+(aq)+6e-→2Cr3+(aq)+7H2O(l) 1.36 Cl2(g)+2e-→2Cl-(aq) 1.358 O2(g)+4H+(aq)+4e-→2H2O(l) 1.229 MnO2(s)+4H+(aq)+2e-→Mn2+(aq)+2H2O(l) 1.224 2IO3-(aq)+12H+(aq)+10e-→I2(s)+6H2O(l) 1.195 Br2(l)+2e-→2Br-(aq) 1.066 VO2+(aq)+2H+(aq)+e-→VO2+(aq)+H2O(l) 0.991 HNO2(aq)+H+(aq)+e-→NO(g)+H2O(l) 0.983 NO3-(aq)+4H+(aq)+3e-→NO(g)+2H2O(l) 0.957 2Hg2+(aq)+2e-→Hg22+(aq) 0.92 Hg2+(aq)+2e-→Hg(l) 0.851 Ag+(aq)+e-→Ag(s) 0.800 Hg22+(aq)+2e-→2Hg(l) 0.797 Fe3+(aq)+e-→Fe2+(aq) 0.771 O2(g)+2H+(aq)+2e-→H2O2(aq) 0.695 MnO4-(aq)+2H2O(l)+3e-→MnO2(s)+4OH-(aq) 0.595 I2(s)+2e-→2I-(aq) 0.536 Cu+(aq)+e-→Cu(s) 0.521 O2(g)+2H2O(l)+4e-→4OH-(aq)…
There are 1.699 x 1022 atoms in 1.000 g of chlorine. Assume that chlorine atoms are spheres of radius 0.099 nm and that they are lined up side by side in a 0.50 g sample. How many miles in length is the line of chlorine atoms in the sample? 1 mi = 1609 m
1 A)5.63,4.42,2.50,or 1.25 1 B)L,g,mol,or kg 2 A)129,0.109,86.0,or 57.5 2 B)g,mol,L,or mg
Bromine is used to prepare the pesticide methyl bromide andflame retardants for plastic electronic housings. It is recoveredfrom seawater, underground brines, and the Dead Sea. The aver-age concentrations of bromine in seawater (d=1.024 g/mL) andthe Dead Sea (d=1.22 g/mL) are 0.065 g/L and 0.50 g/L, re-spectively. What is the mass ratio of bromine in the Dead Sea tothat in seawater?