Please answer all of the questions. I need help so bad. I will rate. Thank you f

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Please answer all of the questions. I need help so bad. I will rate. Thank you for your time.

Write a balanced chemical equation (smallest integer coefficients possible) for the reaction between an acid and a base that leads to the production of Cu(CH Coo)(aq). Use the pull down boxes for states Give the names of the acid, the base, and the salt acid name base name: salt name: To determine the concentration of a solution of hydrefluoric acid, a 100.0-mt sample is placed in a flask and titraned with a 0.1003 M solution of potassium hydroxide. A rolume of 36.40 mL is required to reach the phenolphthalein endpoint. Calculate the concentration of hydrofluoric acid in the original sample. Potassium permangamate in acidic solution is used to titrate a solution of hydrogen peroxide, with which it reacts accoeding o A potassium permanganate solution is prepared by dassolving 16.70 g of KMmO in water and diluting to a total volume of 1.000L. A total of 17.08 mL of this solution is required to reach the enapoint in a titration of a 200.0-mL sample containing H()Determine the concentration of H0, in the original solution A student has 460.0 mL of a 0.1 41S M aqueous solution of Na-C03 to use·n an experiment He accidentally eaves the contarer uncovered and comes back the next week to and only a solid residue. The mass of the residue is 18.66g Determine the chemical formula of this residue. If the sabstance is a drane, ne a period instead of a dot in its foe

Explanation / Answer

1)

Cu(OH)2(s)+2CH3COOH(aq)<---->Cu(CH3COO)2(aq)+ 2H2O(l)

Acid Name : Acetic acid

Base Name : Copper(II)hydroxide

Salt Name : Copper(II)acetate

2)

The reaction between KOH and HF is

HF + KOH - - - - > KF + H2O

this is 1:1reaction

So,

V1×M1 = V2×M2

M1 = V2 ×M2/V1

= 36.40ml×0.1003M/100ml

= 0.0365M

Therefore,

Concentration of HF = 0.0365M

3)

No of mole of KMnO4 = 16.70g/158.032g/mol =0.10567moles

Volume of KMnO4 consumed = 17.08ml

No of moles of KMnO4 consumed = (0.10567mol/1000ml)×17.08ml =0.001805

stoichiometrically, 2moles of KMnO4 react with 5moles of H2O2

Therefore,

No of moles of H2O2 = (5/2)×0.001805mol =0.0045125moles

Volume of H2O2 solution = 200ml

Concentration of H2O2 = (0.0045125mol/200ml)×1000ml = 0.0226M

4)

No of moles of Na2CO3 = (0.1418mol/1000ml)×460ml =0.065228

Mass of Na2CO3 = 105.99g/mol ×0.065228mol=6.9135g

weight of residue =18.66g

Mass of water = 18.66g - 6.9135g = 11.7465g

Molar mass of water = 18.02g

No of moles of water = 11.7465g/18.02g/mol =0.65186

ratio of water to Na2CO3 = 0.65186mol/0.065228mol =10

Therefore,

the chemical formula is Na2CO3.10H2O

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