Please answer all of them. Thank you ! Two solutions, initially at 24.60 degree

Question

Please answer all of them. Thank you !

Two solutions, initially at 24.60 degree C, are mixed in a coffee cup calorimeter (C_cal = 15.5 J/degree C). When a 100.0 mL volume of 0.100 M AgNO_3 solution is mixed with a 100.0 mL sample of 0.200 M NaCl solution, the temperature in the calorimeter rises to 25.30 degree C. Determine the Delta H degree_rxn for the reaction as written below. Assume that the density and heat capacity of the solutions is the same as that of water. NaCl (aq) + AgNO_3(aq) rightarrow AgCl(s) + NaNO_3(aq) Delta H degree_rxn = ? -16kJ -250kJ -140kJ -35kJ -69kJ Which reaction below represents the second ionization of Sr? Sr(g) rightarrow Sr^2+(g) + 2e^- Sr^+(g) rightarrow Sr^+ (g) + e^- Sr^2^+ (g)+ e^-rightarrow> Sr^+(g) Sr^+(g) + e^- rightarrow Sr(g) Sr^2+ (g) + 2e^- rightarrow Sr (g) Give the ground state electron configuration for I. [Kr]4d^10 5s^2 5p^5 [Kr]5s^2 5d^10 5p^6 [Kr]4d^10 5p^6 [Kr]5s^2 5p^6 [Ar]3d^10 4s^2 4p^6

Explanation / Answer

Q=m x dTx S

= 200 x 0.7 x 4.18

= 585.2J

Q cal= 15.5 x 0.7 = 10.85J

Total heat = 596J

Limiting reagent is AgNO3

596J is for 0.01 moles so for 1 mole it is 59.6kJ unfortunately this is nor an option in the answers provided so I will go with option E)

2) second ionization is option B)

question asks for second ionization only So the first ionization that is converting Sr o Sr+ is already done So we need to choose the answer that generates Sr2+ from Sr+ by removing one electron

3) A) is the correct option

we have choose the electronic configuration of I The filled noble gas before I is Kr and then you fill the 4d10 and 5s2 and 5p5 as per the number of electrons of I

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