<u>Answer:</u> The 
for HCN (g) in the reaction is 135.1 kJ/mol.
<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. 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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%29%7D%3D135.1kJ)
Hence, the for HCN (g) in the reaction is 135.1 kJ/mol.
Answer:
For example, friction between our shoes and the floor stop us from slipping and friction between tyres and the road stop cars from skidding. Friction is sometimes unhelpful. For example, if you don't lubricate your bike regularly with oil, the friction in the chain and axles increases.
Explanation:
Answer:
To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
Explanation:
Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.
So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles
We know, number of moles of a compound is the ratio of mass to molar mass of that compound.
So, mass of 2 moles of urea = 
Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution
So, option (C) is correct.
