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what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
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What is the pH of the solution that results from mixing the following four solutions together?
(1) 150. mL of 0.250 M NaCl (2)
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I would determine the molarity of each component. moles NaCl = M x L moles HCl = M x L. moles HNO3 =
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what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
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You must like this problem.
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what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
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2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2 mols HNO3 = M x L = ? Use the coefficients to convert mols HNO3
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what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
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See your other post above.
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what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
Top answer:
See your other post above.
Read more.
what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
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To find the pH of the solution, we need to determine the concentration of the resulting solution
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What is the molarity of the initial HNO3? That's 1000 cc x 1.42 g/cc x 0.70 x (1/63) = approx 16 M
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How many mL of a 0.5M Ca(OH)2 solution are needed to neutralize 38 mL of a 2M HNO3 solution?
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ANSWER IS: x = multiply / = divide 38mL HNO3 X 1 L HNO3 x 2M HNO3 x 1 L Ca(OH)2 x 2M Ca(OH)2 x 1000
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How can you prepare 2M of 200 ml HNO3 solution from 63% stock solution of HNO3 which has a density of 1.4 g/ml?
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Calculate the molarity of the stock solution. 1.4 g/mL x 1000 mL x 0.63 x (1/molar mass HNO3) = ? M
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