A 500 ml sample of vinegar density 106 gml was diluted to

Reference no: EM13346157

A 50.0 mL sample of vinegar (density = 1.06 g/mL) was diluted to 500.0 mL. Titration of a 50.0 mL aliquot of this solution required 42.5 mL of 0.0900 M NaOH to reach the end point. Determin the acid content of the vinegar as wt. % acetic acid.

Reference no: EM13346157

Questions Cloud

The hydrogen gas generated is collected over water at : a piece of sodium metal reacts completely with water as follows.2 nas 2 h2ol 2 naohaq h2gthe hydrogen gas generated

If a computer executes instructions to sum the contents : if a computer executes instructions to sum the contents of memory cells 2 and 999 and store the result in cell 0what

At 25 degree c the heat of fusion of al is 106 kjmol and : at 25 degree c the heat of fusion of al is 10.6 kjmol and the heat of sublimation is 326.4 kjmol. what is the heat of

What is the primary growth front what is the secondary : what is the primary growth front? what is the secondary growth front? what happens when they approach each other? what

A 500 ml sample of vinegar density 106 gml was diluted to : a 50.0 ml sample of vinegar density 1.06 gml was diluted to 500.0 ml.nbsp titration of a 50.0 ml aliquot of this

Calculate the molarity of the following : calculate the molarity of the following solutionsa5.6 mol of sodium chloride dissolved to form 25 l of solution.b5.6 g

Assume that the density of the solution is 100 gcm3 : a solution is made by dissolving 25g of nacl in enough water to make 1.00 l of solution. assume that the density of the

The number of women-years of exposure is less than for : the number of women-years of exposure is less than for established methods and all available evidence should be

Consider a one dimensional crystal with a total of six : consider a one dimensional crystal with a total of six atoms each with mass m seperated by a lattice constant of 0.5

User Account

All Pages