.......A + B ==> C + D I.....0.8..0.8...0...0 C.....-x...-x....x...x E...0.8-x.0.8-x..x...x So C = x = 0.6. and 0.8-x = A = ?
HNO3 is a strong acid; i.e., it ionizes 100%. Therefore, (H^+) = (HNO3). HCOOH is a weak acid; i.e., it ionizes only partially. Therefore, (H^+) < (HCOOH). That means that the acidity of the HCOOH solution will be weaker, the (H^+) will be lower, and the pH will be higher. ...
How can we help you without writing your paper for you. This can't be done in a sentence or two.
How advanced is this class? I suppose the beginning class you would expect mols H2SO4 = grams/molar mas = about 0.0510. Two mols H^+/mol H2SO4 = (H^+) = 2*0.0510. However, in reality, the second H in H2SO4 is not a strong acid as is the first one so it contributes only a minim...
Maxy, you need to rework your post. I expect it hasn't been answered because we don't know what you're asking.
You need to work on your post to make it more sensible. As it stands it isn't clear what you are asking.
You want how many mols Ag? That's M x L = 0.05 x 2L = 0.1. mols = grams/molar mass; therefore, grams = mols x molar mass = 0.1 mol x 107.9 g/mol = ? grams Ag metal.
Look up information on the Friedel-Crafts reaction. Here is a good site showing substitution onto an aryl ring. http://en.wikipedia.org/wiki/Friedel%E2%80%93Crafts_reaction
HClO4 is a strong acid;i.e., it ionizes 100%. HClO4 ==> H^+ + ClO4^- Therefore, 0.01M HClO4 = 0.01M (H^+). Then (H^+)(OH^-) = Kw = 1E-14 Solve for OH^-
P1V1 = P2V2
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