WebThe diameter of zinc atom is 2.6 A. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. Solution. Question 38. A certain particle carries 2.5 x 10-16 C of static electric charge. Calculate the number of electrons present in it. Solution. WebDivide the volume of zinc (cm3) removed by the Surface area (cm^2) and determine the thickness of the layer of each washer. Be sure to use consistent units!! Washer A: Surface Area 10.23cm^2 volume 0.7379cm^3 Washer B: Surface Area 9.69cm^2 volume 0.7099cm^3 Washer C: Surface Area 12.08cm^2 volume 0.8227cm^3 Expert Answer
The diameter of zinc atom is 2.6 ∘A . Calculate (a) radius of zinc atom
WebCalculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. Hard Solution Verified by Toppr (i)The diameter of zinc atom is 2.6 Ao=2.6×10 −10m. The radius of Zn atom is 22.6×10 −10=1.3×10 −10m=130×10 −12m=130 pm. WebAdvanced Math questions and answers. Using the diameter of the zinc atom (7.133 grams/cm^3) (please double the radius!!), calculate the thickness of the zinc layer, in … ionized foot spa detox
The size of atoms - GSU
WebObjective: The objective of this experiment was to determine the size of a zinc atom by reacting the galvanized zinc with 6.0m HCl. Procedure: On sheet (simple enough you do not need to repeat it) Data and Calculations: Initial mass of washer= 9.65260g Diameter of washer= 3.8cm Diameter of washer whole= 1.0cm Final mass of washer= 7.9630g 1. Web6.57 g/cm 3 : Heat of fusion: 7.32 ... to magnesium: both elements exhibit only one normal oxidation state (+2), and the Zn 2+ and Mg 2+ ions are of similar size. Zinc is the 24th … WebSep 12, 2013 · atomic mass of zinc = 65.39 g/mol The Attempt at a Solution For part (a) I use the fact that zinc has atomic mass of 65.39 g/mol. So dividing 7,170,000 (g/m^3) by 65.39 (g/mol), you get 109,649.79 (mol/m^3). Then we divide that by 10^6 to get .10964979 mol/cm^3. Then we multiply that by avagadro's number to get 6.603*10^22 atoms/cm^3. on the ball academy glasgow