<span>when adding a few drops of base causes a large increase in pH</span>
Answer:
a. U235
Explanation:
Atomic number 92 is that of Uranium. It naturally occurs in several isotopes:
U-238, U-235, U-234
? is their a picture if it is please send it
Answer:
![\large \boxed{\text{-237.0 kJ/mol}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B-237.0%20kJ%2Fmol%7D%7D)
Explanation:
2NH₄NO₃(s) ⟶ 2N₂(g) + O₂(g) + 4H₂O(g)
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
![\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})](https://tex.z-dn.net/?f=%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%3D%20%5Csum%20%5CDelta_%7B%5Ctext%7Bf%7D%7D%20H%5E%7B%5Ccirc%7D%20%28%5Ctext%7Bproducts%7D%29%20-%20%5Csum%5CDelta_%7B%5Ctext%7Bf%7D%7DH%5E%7B%5Ccirc%7D%20%28%5Ctext%7Breactants%7D%29)
2NH₄NO₃(s) ⟶ 2N₂(g) + O₂(g) + 4H₂O(g)
ΔH°f/kJ·mol⁻¹: -365.1 0 0 -241.8
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & 4\times(-241.8) - 2\times(-365.1)\\& = & -967.2 + 730.2\\& = & \textbf{-237.0 kJ/mol}\\\end{array}\\\text{The enthalpy of reaction is } \large \boxed{\textbf{-237.0 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%204%5Ctimes%28-241.8%29%20-%202%5Ctimes%28-365.1%29%5C%5C%26%20%3D%20%26%20-967.2%20%2B%20730.2%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-237.0%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20of%20reaction%20is%20%7D%20%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-237.0%20kJ%2Fmol%7D%7D)
Answer:
(1) I shifts toward product and II shifts toward reactant.
Explanation:
Increasing the temperature of an endothermic reaction (∆H is positive) shifts the equilibrium position to the right thus favoring product formation.
Increasing the temperature of an exothermic reaction (∆H is negative) shifts the equilibrium position to the left thus favoring the backward reaction.