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A solution contains 1.569mg of CoSO4(155.0g/mol)per mm. Calculate
a) the volume of 0.007840M EDTA needed to titrated a 25.00mL aliquot of this solution
Mass of sulfate = 0.0001569gSulfate = mass/molar massNumber of moles of cobalt = 0.0001569g/155=0.00000101mol/ml25× 0.00000101= 0.00002525mol
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Water is added to 25ml of 0.866M KNO3 solution until the volume of the solution is exactly 500 ml. What is the concentration of the final solution
C=C0V0/V=0.866*25*10-3/500*10-3=0.0433M
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A 0.3259g of (KPH) required 20.15ml of NaOH for neutralization. Calculate the molarity of NaOH
Answer on Question #97262 – Chemistry – Organic ChemistryTask:A 0.3259g of (KPH) required 20.15ml of NaOH for neutralization. Calculate the molarity of NaOH.Solution:"KHP" - potassium hydrogen phthalate (KHC8H4O4, Molar mass...
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23.91 mL of a standard solution of KHP of molarity 0.1454M used to standardize a solution of NaOH giving final volume of 26.31 mL of NaOH at endpoint. Calculate concentration of NaOH
0.1321M
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a 0.2025g sample consiting of only Bacl2 and kcl required 20.25ml 0.1200M AgNo3 solution for the quantitative precipitation of chloride. calculate the %Ba and %k of the sample
The sample requires 20.25 ml of a 0.1200 M AgNO3 solution... ... A Sample Weighing 0.2025 Grams Consists Of Only BaCl2 And KCl
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What volume of 1.5N NaOH is needed to react with 25ml 4.0N HCl?
NaOH + HCl → NaCl + H2OMolar Mass of NaOH = 39.9971.5 × 39.997 = 59.9955Molar Mass of HCl = 36.458 g145.83-36.458 = 109.374ml
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A 0.2025g sample consisting of only BaCl2 and KCl required 20.25ml of 0.1200M AgNO3 solution for the quantitative precipitation of chloride. Calculate the %Ba and %K in the sample
Solution:let x = mass of KCl, y = mass of BaCl2Thus: x + y = 0.2025 g; → x = 0.2025 - yThe molar mass of BaCl2 is 208.23 g mol-1The...
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A 0.2025 Sample consisting of only Bacl2 and Kcl required 20.25ml of 0.1200M Agno3 solution for the quantitative precipitation of chloride. Calculate the %Ba and %K in the sample.
2AgNO3 + BaCl2 = 2AgCl + Ba(NO3)2CM = n/Vn = CM x Vn (AgNO3) = 0.1200 x 0.02025 = 0.00243 moln (BaCl2) = 1/2 x n (AgNO3) = 0.00243/2 =...
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0.25g of acid is titrated with 108M NaOH, if it takes 27.22ml NaOH to reach the endpoint, what is the equivalent weight of the acid
CM = n / Vn = CM x Vn = m / MM = m / nn (NaOH) = 0.108 x 0.02722 = 0.003 molMacid = 0.25 / 0.003 =...
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A 30.84 mL sample of 0.128 M NaOH is needed to reach an endpoint for a titration of a 5.441 gram sample of Vinegar. Calculate the % acetic acid in the vinegar sample. (Show all work)
First calculate the moles of acetic acid, then the weight of acetic acid, then divide by the weight of vinegar. 1. moles of acetic acid is equivalent to moles of...
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