Answer:
The answers are options b , c and d.
Hope this helps.
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
The least electronegative component in the electron transport chain is the Hydrogen ion.
The more electronegative is NAD+
The other component is H2O,
Next are the energy carrier molecules which are the ADP and ATP
And finally, the most electronegative is O2.
Answer: The standard enthalpy of formation of 
is -252.1 kJ/mol.
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactantss%7D%5D)
Putting the values we get :
![\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%5Ctimes%20H_f%7BFe%7D%2B3%5Ctimes%20H_f%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f%7BFe_2O_3%7D%2B3%5Ctimes%20H_f%7BH_2%7D%5D)
![67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]](https://tex.z-dn.net/?f=67.9kJ%3D%5B%282%5Ctimes%200%29%2B%283%5Ctimes%20H_f%7BH_2O%7D%29%5D-%5B%281%5Ctimes%20-824.2kJ%2Fmol%29%2B3%5Ctimes%200kJ%2Fmol%29%5D)
Thus standard enthalpy of formation of is -252.1 kJ/mol.
