An atom has a diameter of 1.50 Å and the nucleus of that atom has a diameter of 5.00 x 10-5 Å. Determine the fraction of the volume of the atom that is taken up by the nucleus. Assume the atom and the nucleus are a sphere. fraction of atomic volume: Calculate the density of a proton, given that the mass of a proton is 1.0073 amu and the diameter of a proton is 1.70 x 10-¹5 m. density: g/cm³

Help 100% 47 T. "ublic Health Ch HSC 258 - Major Projec x MindTap - Cengage Lea X C The Illustration To T =55750828934189288909969212&elSBN=9781305657571&id=D1061392007&nbld=21... * Q Search t Referonces Use the References to access important values if needed for this question. For the following reaction, 50.4 grams of sulfur dioxide are allowed to react with 17.9 grams of water. sulfur dioxide (g) + water (I) sulfurous acid (H2SO3) (g) grams What is the maximum amount of sulfurous acid (H,SO3) that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? Submit Answer

ne File Edit View History Bookmarks Profiles Tab Evolution of the Genome-Cell X Significant Figures Counter E esc X Mathway | Algebra Problem So x A ALEKS-Lara Althawadi - Lear X C www-awu.aleks.com/alekscgi/x/lsl.exe/1o_u-IgNslkr7j8P3jH-IvTqeviKFP6W0cqJcWJdIACROQwyw24GWHInqCmj000aolPZo5oqa7WgLOML1yaFz8E-CEGkqhv... solution Form M Gmail OKINETICS AND EQUILIBRIUM Setting up a reaction table initial change equilibrium G hess law organic c.... YouTube Maps Suppose a 250. mL flask is filled with 0.80 mol of H₂ and 0.90 mol of NH₂. This reaction becomes possible: N₂(g) + 3H₂(g)2NH, (g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of N₂. You can leave out the M symbol for molarity. Explanation CA 17,306 N₂ 0 X 0 Check 3 Tab Window Help H₂ 0 0 9 NH, 0 0 0…

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