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What is the reason for why a sand bath was used instead of a water bath for this particular pressure reaction?
(a) Instructor preference (either could have been used with equal success).
(b) The chemicals inside of the pressure tube needed to be kept anhydrous (water-free environment) in order for the reaction to proceed.
the final step in the mechanism for the base promoted hydrolysis (saponification) of an ester is a deprotonation of a carboxylic acid by sodium hydroxide to give a carboxylate salt (RCO2Na)
Calculate the molar concentration of uncomplexed Zn2+ in a solution that contains 0.20 mole of Zn(NH3)42+ per liter and 0.0116 M NH3 at equilibrium.
He finds that the total pressure at equilibrium is 40% greater than it was originally. What is Kc for this reaction?
Nitrogen (N2) reacts with hydrogen to give ammonia (NH3): N2 + 3H2 -> 2NH3 If 30 g of N2 reacts with excess hydrogen, what the mass of ammonia formed would be?
A mixture of 82.49 g of aluminum (MM = 26.98 g/mol) and 117.65 g of oxygen (MM = 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.
one hundred grams of water is saturated with NH4Cl at 50 degrees celcius. According to table G if the temperature is lowered to 10 degrees celcuis what is the total amount of NH4Cl that will precipitate.
If you dilute 25.0 of the stock solution to a final volume of 0.460 , what will be the concentration of the diluted solution?
The enthalpy of combustion of methane, Ch4, is -890 KJ/mol. If the gas flow rate of the methane to the burner of a stove is 5 L/min at 25 degrees celsius and assuming complete heat transfer from combustion
pH = log[H3O+] , pH + pOH = 14, [H3O+] * [OH-] = 1.0 X 10^-14, pH = -log(0.001) = 3 Could you show me a few of the steps between line one and two in both of these conclusions?
A 4.30 g nugget of pure gold absorbed 293 J of heat. The initial temperature was 29.0°C. What was the final temperature?
Calculate the pH of the titration mixture after 10.0, 20.0 and 30.0 mL of base have been added. (the ka for acetic acid is 1.76 x 10^-5).
A student has prepared a 0.309 M solution of Barium hydroxide and a 0.443M solution of hydrochloric acid for you. Determine the volume of the hydrochloric acid solution will be needed to exactly neutralize 10.0 mL of the Barium chloride solution.
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