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125 mL of 1.00 HCl with 9.00 g of KOH. if the yield of KCL is 7.51 g what is the percent yield of the reaction
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125 mL of 1.00 M HCl with 9.00 g of KOH. if the yield of KCL is 7.51 g what is the percent yield of the reaction
HCl + KOH -> KCl
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The procedure in this example will work the problem for you.
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1. What is the molarity of an HCl solution if 43.6 mL of a 0.125 M KOH solution are needed to titrate a 25.0 mL sample of the
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The balanced chemical equation shows a 1:1 mole ratio between KOH and HCl. This means that the moles
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Which of these pairs of substances in aqueous solution would constitute a buffer?
a. HCl + KCl b. KHPO4 + K2HPO4 c. KOH + KCl d.
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A buffer consists of a. a weak acid and a salt of the weak acid. b. a weak base and a salt of the
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Which of these pairs of substances in aqueous solution would constitute a buffer?
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Buffers are mixtures of weak acid and a salt of the weak acid OR a weak base with a salt of the weak
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in which of the following pairs of solutions ,where thre is no effect on the PH after dilution?
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9.7g ofamixture of KOH and KCL was to
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NOTE: You mean 20 mL or 20 cc or 20 cm^3 and not 20 cm. KOH + HCl ==> KCl + H2O millimols HCl = mL x
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0.5 L of a 0.30 M HCl solution is titrated with a solution of 0.6 M KOH.
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The secret to working these problems is to know where you are on the titration curve and solve
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How would you find the pH of a solution prepared by mixing 50 mL of 0.125 M KOH with o.050 L of 0.125 M HCl?
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mols KOH = M x L = ? mols HCl = M x L = ? Determine which is in excess, the pH = -log(H^+) if HCl is
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9. According to Arrhenius, which of the following groups contain:
i)only acids ii)only bases a. NaOH, H2CO3, KCl b. MgCl2, H2SO4,
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The two comments above me have no lives and are useless to this world. Parents should have disowned
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The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is __________.
i know the answer is
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Hi laney, For this question, it would be appropriate for you to ask yourself.. what is present in
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