Umm its not a i think... lemme see lol
The temperature is -8 F
Answer D
Answer:
![\Delta _RH=4x10^{-3}\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_RH%3D4x10%5E%7B-3%7D%5Cfrac%7BkJ%7D%7Bmol%7D)
Explanation:
Hello,
In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:
![\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_RH%3D%5Cfrac%7B12J%7D%7B3mol%7D%20%3D4%5Cfrac%7BJ%7D%7Bmol%7D)
Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:
![\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_RH%3D4%5Cfrac%7BJ%7D%7Bmol%7D%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D4x10%5E%7B-3%7D%5Cfrac%7BkJ%7D%7Bmol%7D)
Best regards.
5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).
We can calculate the moles of CaC₂O₄ obtained considering the following relationships.
- The molar mass of K₂C₂O₄ is 184.24 g/mol.
- The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.
![5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4](https://tex.z-dn.net/?f=5%20g%20K_2C_2O_4%20%5Ctimes%20%5Cfrac%7B1molK_2C_2O_4%7D%7B184.24gK_2C_2O_4%7D%20%20%5Ctimes%20%5Cfrac%7B1molCaC_2O_4%7D%7B2molK_2C_2O_4%7D%20%3D%200.03molCaC_2O_4)
5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Learn more: brainly.com/question/15288923