Answer:
![2.4*10^{7}cm^{3}](https://tex.z-dn.net/?f=2.4%2A10%5E%7B7%7Dcm%5E%7B3%7D)
Explanation:
First you should calculate the volume of a big sphere,so:
![V_{big}=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V_%7Bbig%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
![V_{big}=\frac{4}{3}\pi (181cm)^{3}](https://tex.z-dn.net/?f=V_%7Bbig%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%28181cm%29%5E%7B3%7D)
![V_{big}=2.4*10^{7}cm^{3}](https://tex.z-dn.net/?f=V_%7Bbig%7D%3D2.4%2A10%5E%7B7%7Dcm%5E%7B3%7D)
Then you calculate the volume of a small spehre, so:
![V_{small}=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V_%7Bsmall%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
![V_{small}=\frac{4}{3}\pi (5.01cm)^{3}](https://tex.z-dn.net/?f=V_%7Bsmall%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%285.01cm%29%5E%7B3%7D)
![V_{small}=5.3*10^{2}cm^{3}](https://tex.z-dn.net/?f=V_%7Bsmall%7D%3D5.3%2A10%5E%7B2%7Dcm%5E%7B3%7D)
Finally you subtract the two quantities:
![V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}](https://tex.z-dn.net/?f=V_%7Bbig%7D-V_%7Bsmall%7D%3D2.4%2A10%5E%7B7%7Dcm%5E%7B3%7D-5.3%2A10%5E%7B2%7Dcm%5E%7B3%7D)
![V_{big}-V_{small}=2.4*10^{7}cm^{3}](https://tex.z-dn.net/?f=V_%7Bbig%7D-V_%7Bsmall%7D%3D2.4%2A10%5E%7B7%7Dcm%5E%7B3%7D)
Answer: ![\Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D-20kJ%2Fmol-%28%2B66kJ%2Fmol%29)
Explanation:
Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.
It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.
![\Delta H=H_{products}-H_{reactants}](https://tex.z-dn.net/?f=%5CDelta%20H%3DH_%7Bproducts%7D-H_%7Breactants%7D)
= enthalpy change = ?
= enthalpy of products
= enthalpy of reactants
For the given reaction :
![2NO_2(g)\rightarrow N_2O_4(l)](https://tex.z-dn.net/?f=2NO_2%28g%29%5Crightarrow%20N_2O_4%28l%29)
![\Delta H=H_{N_2O_4}-2\times H_{NO_2}](https://tex.z-dn.net/?f=%5CDelta%20H%3DH_%7BN_2O_4%7D-2%5Ctimes%20H_%7BNO_2%7D)
![\Delta H=-20kJ/mol-(+66kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20H%3D-20kJ%2Fmol-%28%2B66kJ%2Fmol%29)
15.6gC3H8*44.1gC3H8*32gO2=11.32
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
brainly.com/question/15289741
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