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You have an unlimited amount of 12M HCl stock solution. You need to neutralize 2.00 grams of NaOH by slowly adding 0.100M HCl solution.
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You have an unlimited amount of 12M HCl stock solution. You need to neutralize 2.00 grams of NaOH by slowly adding 0.100M HCl
Top answer:
2.00 g = 2.00/molar mass NaOH = about 0.05 moles NaOH and that will take 0.05 moles HCl. mL12M x
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Please help me with these! Thanks :)
Describe, in detail, how to make 5.5 liters of a 2.5 molar HCl solution from a 10.5 molar
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you are diluting it 10.5/2.5 times, or 4.2 times. Which means, you want one part stock solution, and
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You have 14.0mL of a 0.12M NaOH solution. How many moles of HCl will it take to neutralize this solution?
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same as # mols NaOH mols NaOH = M x L
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do it as a proportion: x/.265g = 1E-6L/1E-3L x= .265 mg x=.000265grams check my thinking.
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density 1.191 g/mL. mass 1000 mL = 1.191 g/mL x 1000 mL = 1191 g. mass HCl in that 1L = 1191 x 0.80
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how many moles of HCL is 2.5 grams? The balanced equation states it will take half that number of
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Describe in detail by words how to prepare 200ml of .150M HCl from a 1.00 M HCl stock solution? Need help with this please?
i
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Thanks for providing the numbers. I don't know what you did wrong but here is how you do it. Pay
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Prepare a 0.0200 mM dye solution at low pH by dilution 0.40 ml of the stock 0.100 mM chemical X into the appropriate volume of
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Calculate the volume of 2.503 N HNO3 added to 100 mL of HCl in order to prepare 1.1251 N acid solution from the following data:
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To calculate the volume of 2.503 N HNO3 needed to prepare a 1.1251 N acid solution, we can follow
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1. how many grams of KOH will neutralize (react with) 0.15 mol HNO3
2. how many liters of 1 M HCl will neutralize 10 ML of 9.2 M
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KOH + HNO3 ==> KNO3 + H2O so you know it's 1 mol KOH to 1 mol HNO3. mols HNO3 = M x L = ? mols KOH =
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