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A beaker contains 50.0 mL of 0.952 M solution of a diprotic acid. How many grams of Mg(OH)2 are required to neutralize this solution?

What do you want? I gave you step by step instructions at your first post just a half dozen or so posts below. All you need to do is to substitute the numbers.

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A beaker contains 50.0 mL pf 0.953 M solution of a diprotic acid. How many grams of Mg(OH)2 are required to neutralize this

Top answer: Mg(OH)2 + H2P ==> MgP + 2H2O moles H2P = M x L = ?? Since 1 mole Mg(OH)2 = 1 mole H2P, then mols Read more.
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Top answer: I think that's no and yes but here is what you do. All of these are done stepwise and most follow Read more.
Hi, how do I calculate how many grams of unknown diprotic acid are needed to make 150mL of 0.750N (normal) solution?and to make

Top answer: a). How many equivalents do you need? THat is L x N = 0.150 x 0.750 = ? grams = equivalent weight x Read more.
A beaker contains 100 grams of 1.20 M NaCl. If you transfer 50% of the solution to another beaker, what is the molarity of the

Top answer: Think about it a little. Suppose we have a beaker containing a 1 M solution of NaCl. Suppose the Read more.
G) The mass of a beaker is 40.94 g. After 5.00 mL of concentrated hydrochloric acid solution is pipetted into the beaker, the

Top answer: 46.92 g = mass beaker + HCl -40.94 g = mass beaker ------------------ ....5.98 = mass HCl density = Read more.
You have two beakers of acid solution. One beaker contains a 17% acid solution, and the other beaker contains a 91% ACID

Top answer: amount of 17% solution --- x L amount of 91% solution --- 2-x L .17x + .91(2-x) = .56(2) solve for x Read more.
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1. Harry titrates vinegar with 0.952 M NaOH. The sample was 25.75 mL of vinegar, and it took 19.85 ml of 0.952 M NaOH to reach

Top answer: millimoles NaOH = mL x M = 19.85 x 0.952 = ? = estimated 19; therefore, estimated 19 millimols Read more.
A chemist had a 1.5L solution of hydrochloric acid in a beaker. The chemist removed 25mL from the beaker. How many milliliters

Top answer: You may want to start by typing all of the problem. For the solution part of the problem, you Read more.