Answer:
The option D is correct answer.
This is an incomplete question, here is a complete question.
A 6.55 g sample of aniline
molar mass = 93.13 g/mol) was combusted in a bomb calorimeter. If the temperature rose by 32.9°C, use the information below to determine the heat capacity of the calorimeter.
![4C_6H_5NH_2(l)+35O_2(g)\rightarrow 24CO_2(g)+14H_2O(g)+4NO_2(g)](https://tex.z-dn.net/?f=4C_6H_5NH_2%28l%29%2B35O_2%28g%29%5Crightarrow%2024CO_2%28g%29%2B14H_2O%28g%29%2B4NO_2%28g%29)
ΔH°rxn = -1.28 × 10⁴ kJ
Answer : The heat capacity of the calorimeter is, ![6.84kJ/^oC](https://tex.z-dn.net/?f=6.84kJ%2F%5EoC)
Explanation :
First we have to calculate the moles of aniline.
![\text{Moles of aniline}=\frac{\text{Mass of aniline}}{\text{Molar mass of aniline}}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20aniline%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20aniline%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20aniline%7D%7D)
![\text{Moles of aniline}=\frac{6.55g}{93.13g/mol}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20aniline%7D%3D%5Cfrac%7B6.55g%7D%7B93.13g%2Fmol%7D)
![\text{Moles of aniline}=0.0703mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20aniline%7D%3D0.0703mol)
Now we have to calculate the heat releases.
As, 4 mole of aniline on combustion to releases heat = ![1.28\times 10^4kJ](https://tex.z-dn.net/?f=1.28%5Ctimes%2010%5E4kJ)
So, 0.0703 mole of aniline on combustion to releases heat = ![\frac{0.0703}{4}\times 1.28\times 10^4kJ=224.96kJ](https://tex.z-dn.net/?f=%5Cfrac%7B0.0703%7D%7B4%7D%5Ctimes%201.28%5Ctimes%2010%5E4kJ%3D224.96kJ)
Now we have to calculate the heat capacity of the calorimeter.
![\text{Heat capacity of the calorimeter}=\frac{\text{Heat releases}}{\text{Change in temperature}}](https://tex.z-dn.net/?f=%5Ctext%7BHeat%20capacity%20of%20the%20calorimeter%7D%3D%5Cfrac%7B%5Ctext%7BHeat%20releases%7D%7D%7B%5Ctext%7BChange%20in%20temperature%7D%7D)
![\text{Heat capacity of the calorimeter}=\frac{224.96kJ}{32.9^oC}](https://tex.z-dn.net/?f=%5Ctext%7BHeat%20capacity%20of%20the%20calorimeter%7D%3D%5Cfrac%7B224.96kJ%7D%7B32.9%5EoC%7D)
![\text{Heat capacity of the calorimeter}=6.84kJ/^oC](https://tex.z-dn.net/?f=%5Ctext%7BHeat%20capacity%20of%20the%20calorimeter%7D%3D6.84kJ%2F%5EoC)
Thus, the heat capacity of the calorimeter is, ![6.84kJ/^oC](https://tex.z-dn.net/?f=6.84kJ%2F%5EoC)
Molarity is defined as the number of moles in 1 L of solution.
Mass of NaNO₃ present - 85.0 g
Molar mass of NaNO₃ - 85 g/mol
The number of NaNO₃ moles - 85.0 g / 85 g/mol = 1 mol
Volume of solution - 750 mL
The number of moles in 0.750 mL - 1 mol
Therefore in 1 L solution, number of NaNO₃ moles - 1/0.75 = 1.33 mol
Therefore molarity of solution - 1.33 M
I’m sorry I’m really not sure how to answer this but have to respond to have this app I hope you have a good day:)
Answer:
The larger the species, the further away from the nucleus the outer electrons are and hence the less strongly they are held. This means the bigger atoms will be more polarizable. So, the order from biggest-smallest molecules (strongest to weakest IMF) is iodine, bromine and finally chlorine.
Explanation: