Answer:
The answer is A.
Explanation:
Gas particles do not have a fixed position or volume so they move in <em>r</em><em>a</em><em>n</em><em>d</em><em>o</em><em>m</em><em> </em><em>s</em><em>p</em><em>e</em><em>e</em><em>d</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>d</em><em>i</em><em>r</em><em>e</em><em>c</em><em>t</em><em>i</em><em>o</em><em>n</em><em>s</em>.
(Correct me if I am wrong)
<span>7.15 degrees C
The specific heat capacity of water is 4.1813 J/(g*K). So we have 3 values with the units kJ, g and J/(g*K). We can trivially convert from kJ to J by multiplying by 1000. And we want to get a result with the unit K (degrees Kelvin). So let's do it. First, let's cancel out the g unit by multiplying.
4.1813 J/(g*K) * 485 g = 2027.9305 J/K
Now we can cancel out the J unit by dividing. But if we divide by the energy, we'll be left with the reciprocal of K, not K. So instead divide by the J/K unit. So
14500 J / 2027.9305 J/K = 7.150146418 K
Rounding to 3 significant figures gives us 7.15 K.
And since degrees C and degrees K are the same size, the temperature will increase by 7.15 degrees C</span>
<u>Answer:</u> The enthalpy of the reaction is coming out to be -902 kJ.
<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_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%284%5Ctimes%20%5CDelta%20H_f_%7B%28NO%28g%29%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%284%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%284%5Ctimes%20%2891.3%29%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%284%5Ctimes%20%28-45.9%29%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-902kJ)
Hence, the enthalpy of the reaction is coming out to be -902 kJ.
The answer is to heat the water up to increase the rate that which it dissolves
