That happens because in oil and vinegar have a certain density weight.
-A<span>n object will float if it is less dense than the liquid it is placed in.
<span>-An object will sink if it is more dense than the liquid it is placed in.</span></span>
Answer:
Here's what I get
Explanation:
From the scattering experiments, Rutherford concluded that the positive charge was concentrated in the nucleus of the atom.
This contradicted the plum pudding model, which assumed that the positive charge was uniformly distributed throughout the atom.
<u>Answer:</u> The enthalpy of the formation of
is coming out to be -410.8 kJ/mol.Z
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol](https://tex.z-dn.net/?f=81.1%3D%5B%281%5Ctimes%20%28226.7%29%7D%29%2B%284%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%3D-410.8kJ%2Fmol)
Hence, the enthalpy of the formation of
is coming out to be -410.8 kJ/mol.
Idk but I’ll try to answer