Chemistry Experiment, science homework help

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Kindly answer the questions for the chemistry experiment data provided.

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Analysis of an Unknown Solid Acid DATA AND REPORT SHEET 82 1. Unknown number Mol KHP Mass KHP Unknown Sample %KHP 1 Mass of Initial Final Volume buret buret NaOH reading reading used .6300g on 13.213.20 1.647-gom 14.11 14.10L 1.6385 om 14.1 2 3 2. Average %KHP 3. Standard deviation Questions: 1. Determine the %KHP in an unknown sample (mass = 645 mg) that required an average of 15.27 mL of 0.0988 M NaOH to reach the endpoint. Show calculations. 2. If you prepared a solution of KOH by weighing out 6.255 g of solid KOH that had a purity of 87.5% and then diluted this to a final volume of 1 L in a volumetric flask, what would be the %KHP in an unknown sample (mass = 0.601 g) if it required 21.36 mL of KOH titrant to reach the endpoint? Show calculations. 30 Exp 5 Analysis of an Unknown Solid Acid
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Explanation & Answer

Hi , Please check the attached file for details, let me know if you have any question, thank you. Best, James

KHP+ NaOH- KNaP +H2O
1
X

1
0.0988*0.0138

X = 0.0988*0.0138 = 0.001363 moles
The mass of KHP is
The molar mass of of KHP = 204.2 g/mol
0.001363*204.2 = =0.27841g = 278.41 mg
The % of KHP is then
278.41/630 *100% = 44.19%
Similarly we ...


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