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Questions (2)
C2H4O(g) �¨ CH4(g) + CO(g) Find the rate law of C2H4O?
The following kinetic data were observed for the reaction at 688 K.
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The diving atmosphere used by the U.S. Navy in its undersea Sea-Lab experiments consisted of 0.036 mole fraction O2 and 0.058
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Answers (4)
6.1 s-
it is 6.1 second inverse (s-)
Step 1 - molar mass Na2CO3 = 106 g/mol Step 2 - molar mass Na2CO3 *10 H2O = 106 + ( 18 x 10) = 286 g/mol Step 3 - moles Na2CO3*10H2O = 14.8 / 286 = 0.0517 Step 4 - moles water = 10 x 0.0517 = 0.517 Step 5 - mass water = 18 x 0.517 = 9.31 g Step 6 - Total
CaCl2 = have 3 ions delta temp = 3*(-1.858)molality = -5.57 x molality molality = moles solute/kg solvent. assume 0.75 kg solvent, then mass of solute must be 250 grams, to give a 25 %. moles CaCl2 = 250/110 = 2.27 mol molality = 2.27 mol / 0.75 kg = 3.03
