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e. 45.0 14) Calculate the work energy, w, gained or lost by the system when ags

ID: 590860 • Letter: E

Question

e. 45.0 14) Calculate the work energy, w, gained or lost by the system when ags constant external pressure of 1.5 am(IL, atm-101 expands fromm 15 L to 35 L agit a. +5.3 kJ b. +3.0 k 5.3 kJ d. -3.0kJ 15) 1r 4.89 g of ZnClh is dissolved in enough water to give a total volume of 500 ml, what is the molarity of the solution? a 0.217 M b. 1.33 MM c. 0.0179 M d. 0.849 M e. 0.0717 M 16) Give the complete ionic equation for the reaction Gf any) that occurs when aquecous solutions of ihium and copper (II) nitrate are mixed 2Li+(ag) + S2-(aq) + Cu2+(ag) + 2 Nor(ag) Cu2+(ag) + S2-(ag) + 2 LiNOya) b. a. 2Li+(aq) + S2Yag) + Cu2+(aq) + 2 NO3Yag) Casa) + 2 Li+(ag) + 2 NO3-(ag) Li+(aq) + s-(aq) + Cu+ (aq) + NO3-(aq-CuS(s) + LiNO3faq) c. e No reaction occurs 17) Determine the oxidizing agent in the following reaction Ni(s) + 2 AgCIO4(aq) Ni(CIO4)2(aq) + 2 Ags) Ni(s) This is not an oxidation-reduction reaction. d. e· 18) How much heat is absorbed when 45.00 g of C(s) reacts in the presence of excess SO2(g) to produce CS2 CO(g) according to the following chemical equation? 5C(s) + 2 SO2(g) CS2() + 4 CO(g) = 2399 kJ a. 2158 k b. 239.9 k c. 898.5 kJ d. 179.8 k

Explanation / Answer

14)

W = -Pext(Vf-Vi)

P = 1.5atm

Vi = 15.0 L

Vf = 35.0 L

put values in above expression

W = -Pext(Vf-Vi)

W = -1.5atm (35.0L - 15.0L)

W = -30.0 atm.L

W = -30.0 *101.4 J

W = -3040 J

W = -3.0 KJ

Answer: d

15)

Molar mass of ZnCl2 = 1*MM(Zn) + 2*MM(Cl)

= 1*65.38 + 2*35.45

= 136.28 g/mol

mass of ZnCl2 = 4.89 g

we have below equation to be used:

number of mol of ZnCl2,

n = mass of ZnCl2/molar mass of ZnCl2

=(4.89 g)/(136.28 g/mol)

= 3.588*10^-2 mol

volume , V = 500 mL

= 0.5 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 3.588*10^-2/0.5

= 7.17*10^-2 M

Molar mass of ZnCl2 = 1*MM(Zn) + 2*MM(Cl)

= 1*65.38 + 2*35.45

= 136.28 g/mol

mass of ZnCl2 = 4.89 g

we have below equation to be used:

number of mol of ZnCl2,

n = mass of ZnCl2/molar mass of ZnCl2

=(4.89 g)/(136.28 g/mol)

= 3.588*10^-2 mol

volume , V = 500 mL

= 0.5 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 3.588*10^-2/0.5

= 7.176*10^-2 M

= 0.0717 M

Answer: e

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