B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.
Answer:
Explanation:
N₂ + 3H₂ = 2 NH₃
1 vol 2 vol
786 liters 1572 liters
786 liters of dinitrogen will result in the production of 1572 liters of ammonia
volume of ammonia V₁ = 1572 liters
temperature T₁ = 222 + 273 = 495 K
pressure = .35 atm
We shall find this volume at NTP
volume V₂ = ?
pressure = 1 atm
temperature T₂ = 273


liter .
mol weight of ammonia = 17
At NTP mass of 22.4 liter of ammonia will have mass of 17 gm
mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm
= 230.28 gm
=.23 kg / sec .
Rate of production of ammonia = .23 kg /s .
Some of the particles undergo a phase change and become a gas due to burning. An example would be CO2, which is lost in the air around/escapes into the atmosphere. Hope that helps!
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.
Let the empirical formula of compound be
.
The ratio of number of moles of C, H and O can be calculated as follows:

Simplifying the ratio,

Thus, the value of x, y and z will be 1, 2 and 1 respectively.
Therefore, the empirical formula will be
.