Answer:
see below
Explanation:
for A + 2B => Products ...
Rate Law => Rate =k[A][B]ˣ
As shown in expression, A & B are included, C is not.
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
The change in entropy of the surrounding is -146.11 J/K.
Explanation:
Enthalpy of formation of iodine gas = 
Enthalpy of formation of chlorine gas = 
Enthalpy of formation of ICl gas = 
The equation used to calculate enthalpy change is of a reaction is:
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28ICl%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28I_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Cl_2%29%7D%29%5D)
![=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol](https://tex.z-dn.net/?f=%3D%5B2%5Ctimes%2017.78%20kJ%2Fmol%5D-%5B1%5Ctimes%200%20kJ%2Fmol%2B1%5Ctimes%2062.436%20kJ%2Fmol%5D%3D-26.878%20kJ%2Fmol)
Enthaply change when 1.62 moles of iodine gas recast:

Entropy of the surrounding = 

1 kJ = 1000 J
The change in entropy of the surrounding is -146.11 J/K.
Answer:
Weather, environment, ecosystem, species, rain forests
Explanation: