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What would be the result if you added an amount of NaOH that exceeded the amount of your weak acid in the diluted acid solution? Show how you would calculate the pH of the solution if you added 20.0 mL of 0.50M NaOH to 8.0mL of 1.0M of your weak acid. Show all work.
A 15.0 grams lead ball at 25.0 celcius ws heated with 40.5 Joules of heat. Given the specifc heat of lead is 0.128 J/ grams. celsius, what is the final temperature of the lead?
4hclg o2g ---gt 2h2og 2cl2ga. if the rate that hydrogen chloride is being lost is 0.10 ms what is the rate that
Write the balanced net ionic equation for the reaction
If 77.0 mL of hydrogen gas is collected over water at 50.0 degrees celsuis and 763 mm Hg, what is the volume of dry gas at standard conditions? The vapor pressure of water is 92.5mm Hg at 50.0 degrees celsuis .
Water in the sample extracts into the methanol, which is then analyzed. If GC analysis of the methanol yields a peak height of 13.66, what is the % w/w water in the antibotic?
Substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can activate or deactivate the ring to substitution
Determine the number of cells per milliliter of the original culture when the total number of colonies and the total number dilutions are as follows
Glucose makes up about 0.10 % by mass of human blood. how do you calculate the molality.What further information would you need to determine the molarity of the solution
Compute pH of a solution containing .030 moles of TRIS and .020 moles of the salt TRISHCl in a total volume of 1.50L
two moles of an ideals gas at 500k and 10 atm pressure are contained in an insulated piston-cylinder arrangement. the
A 680 MW coal fired power plant is 34% efficient at producing electricity. The coal has a heating value of 20,000 kJ/kg, an ash content of 5.5% and a sulfur content of 4.5%. Note: 1 MW = 1000 kW , and 1 kW = 1 kJ/s. a. What is the daily input rate..
What is the pH of the solution that results from adding 30mL of 0.015M KOH to 50mL of 0.015M benzoic acid, Ka = 6.3 x 10^-5
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