Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
<u>Answer:</u> For the given equation, only iron has the value of 
equal to 0 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^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_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of equal to 0 kJ.
Separation will be achieved if one component adheres to the stationary phase more than the other component does.
Physical change alters a given material without changing its chemical.
Answer:
818.2 g.
Explanation:
- Molarity is the no. of moles of solute per 1.0 L of the solution.
<em>M = (no. of moles of NaCl)/(Volume of the solution (L))</em>
<em></em>
M = 2.0 M.
no. of moles of NaCl = ??? mol,
Volume of the solution = 7.0 L.
∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)
∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.
- To find the mass of NaCl, we can use the relation:
<em>no. of moles of NaCl = mass/molar mass</em>
<em></em>
<em>∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) =</em> (14.0 mol)*(58.44 g/mol) = <em>818.2 g.</em>
