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The amount of l3- in a solution can be determined by titration with a solution containing a known concentration of s2o32-(aq) (thiosulfate ion). The determination is based on the net ionic equation Given that it requires 38.2 mL of 0.450 M Na2s2o3(aq) to titrate a 25.0-ml sample of l0-(aq), calculate the molarity of l0-(aq) in the solution. The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.99-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCI(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+(ag). The Sb3+(aq) is completely oxidized by 39.4 mL of a 0.115 M aqueous solution of KBr03(aq). The unbalanced equation for the reaction is calculate the amount of antimony in the sample and its percentage in the ore. Write a balanced overall reaction from these unbalanced half-reactions. Cu rightarrow cu2+ Ag rightarrow Ag balanced overall reaction: An experiment was performed to test the relative activities of four metals, A, B, C, and D. Each solid metal was mixed with solutions of each of the other cations as shown in this chart. The results were indicated as follows y=yes a reaction occurred N=no a reaction did not occur For example, when A+ (aq) was mixed with B (s), a reaction occurred. Use this data to rank the neutral metals according to their activity. Rank these species by their ability to act as an oxidizing agent. Best oxidizing agent Best reducing agent. Best reducing agent Mg2+ Cu2t Ct2* Ag+ For the following reaction kcio3 rightarrow kci+3/2o2 assign oxidation states to each element on each side of the equation Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. Phases are optional. H2SO4 + 2KOH rightarrow 2H2O+K2SO4 0.950 L of 0.450 M H2SO4 is mixed with 0.900 L of 0.290 M KOH. What concentration of sulfuric acid remains after neutralization? Enter the net ionic equation, including phases, for the reaction of AgNo3(aq) with BaO(aq). Write the net ionic equation (including phases) that corresponds to Fe(No3)2(aq)+Na2S(aq) rightarrow FeS(s) + 2NaNO3 (aq) If a solution containing 126.27 g of mercury(ll) chlorate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction? Write the balanced equation for the reaction of aqueous Pb(CI03)2 with aqueous Nal. Include phases. What mass of precipitate will form if 1.50 L of concentrated Pb(CI03)2 is mixed with 0.850 L of 0.220 M Nal? Assume the reaction goes to completion. Predict whether the following compounds are soluble or insoluble in water: CuC03 KN03 Pb(N03)2 BaS04 PbCI2 CuCI2

Explanation / Answer

1) given 38.2 ml of 0.450 M Na2S203

moles of Na2S203 = 38.2 x 0.450 /1000

moles of Na2S203 = 0.01719

Na2S203 ---> 2Na+ + S2003 -2

So 1 mole of Na2S203 gives 1 mole of S203

So 0.01719 moles of S203 is used

from the equation .

mole of I3- = moles of S203 /2

= 0.01719/2

= 8.595 x 10-3

molariy = moles x 1000/volume

= 8.595 x 10-3 x 1000/ 25

= 0.3438

molarity is 0.3438

2)

3Sb^3+ + BrO3^- + 6H^+ ==> 3Sb^5+ + Br^- + 3H2O

moles of Br03- = 39.4 x 0.115 /1000 = 4.531 x 10-3

so from the reaction ,we know that

1 moles of Br03- requires 3 moles of Sb+3

4.531 x 10-3 moles of Br03- requires x moles of Sb+3

x = 4.531 x 10-3 x 3

x =13.593 x 10-3

amount of antimony = 13.593 x 10-3 x 121.76

amount of antimony = 1.655 g

amount of antimony is 1.655 grams

percentage = 1.655 /7.99 = 20.71

percentage is 20.71 %

3)

4) reactivity order is A>B>C>D ,A is more reactive

5) best reducing agent Mg+2 > Cr+2 > Cu > Ag+

6) a) precipitation

b) redox

c) redox

d) Acid -base

7) K ----> +1 ,+1

Cl -----> +5 ,-1

0 -----> -2 ,0

oxidized is Oxygen and Cl is reduced

8) 2K0H + H2S04 -----> K2S04 + 2H20

moles of KOH = 0.9 x 0.29 = 0.261

moles of H2S04 = 0.95 x 0.45 = 0.4275

1 mole of H2S04 require 2 moles of KOH

S0 x moles of H2S04 require 0.261 moles of KOH

x= 0.261/2 = 0.1305

moles of H2S04 remaining = 0.4275 - 0.1305 = 0.297

volume = 0.9 +0.95 = 1.85

molarity = 0.297 /1.85 = 0.1605

molariy is 0.1605 M

9)

2Ag + (aq) + 2OH-(aq) ---> Ag2O(s) + H2O(l)

10)

Fe2+(aq) + S2- (aq) ---> FeS(s)

11) Hg(Cl03)2 + Na2S ----> HgS + 2NaCl03

moles of Na2S = 17.796 / 78 =0.22815

moles of Hg(Cl03 ) 2 = 126.27 / 367.5 = 0.34359

moles of Hg(Cl03)2 remaining = 0.34359 - 0.22815 = 0.11544

mass of precipitate formed = 0.22815 x 232.5 = 53.045 grams

12)

Pb(Cl03)2 (aq) + 2NaI (aq) ----> PbI2 (s) + 2 NaCl03 (aq)

moles of NaI = 0.850 x 0.22 = 0.187

moles of precipitate = 0.187/2

mass of precipitate = 0.187 x 461 /2 = 43.1035

43.1035 grams of percipitate is formed

13) Soluble - KN03 ,Pb(N03)2 ,CuCl2

Insolube -- CuC03 , BaS04 ,PbCl2

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