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Two acid solutions are to be mixed together. Solution A is 30% acid by volume. Solution B is 70% acid by volume.
The 800
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.30A + .70(800-A) = .54(800)
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I need help figuring out equations for a math problem. This is the question:
For a chemistry experiment Marie needs 300 ml of a
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BUT 0.05x + 0.2y = 0.15 * 300 !!!!!
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What is the mole of oxalic acid?
Equation:H2C2O4 + 2NaOH---> Na2C2O4 + 2H2O *Moles of oxalic acid dihydrate is 0.032 mole. *
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NaOH
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What is the mole of oxalic acid?
Equation:H2C2O4 + 2NaOH---> Na2C2O4 + 2H2O *Moles of oxalic acid dihydrate is 0.032 mole. *
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To determine the number of moles of oxalic acid, we can use the provided information about the moles
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A food scientist also tested some ethanoic acid solution. Ethanoic acid is the acid which is found in vinegar.His results are
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To answer these questions, we need to use the given information about the volume of ethanoic acid
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Hydrobromic acid is a strong acid. If you have 1 L of a 7.5 M solution of Hydrbromic acid?
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1) HBr, being a strong acid, ionizes 100%; therefore, (H^+) = 7.5 < 2) pH = -log(H3O^+) 3) mL x M =
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A solution of acetic acid was prepared by mixing glacial acetic acid with water.The solution was 25% of acetic acid by
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Glacial acetic acid is about 17.5M mL1 x %1 = mL2 x %2 mL1 x 17.5 = 120 mL x 25% mL 1 = ?
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A 30.00 mL volume of a weak acid, HA, (Ka = 3.8 x 10^-6) is titrated with 39.00 mL of 0.0958 M NaOH to the equivalence point.
A.)
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HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) A) pure HA. Set up ICE chart for pure HA, substitute into Ka
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A 1.00M solution of a hypothetical weak acid (HB) has a pH of 2.85. A solution is created using 0.10M of a salt NaB ('B' being
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Use the 1 M HB and pH of 2.85 to determine pKa for HB. The remaining part of the BUFFER problem is
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You wish to make a 0.195 M hydrochloric acid solution from a stock solution of 6.00 M hydrochloric acid. How much concentrated
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mL1 x M1 = mL2 x M2 mL1 x 6.00 = 50.0 mL x 0.195 So you want mL1 of the 6.00 M solution to be placed
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