The reaction occurs in a similar way as magnesium does, but much less vigorous. Strong heating is required to make iron powder burn in oxygen. The reaction gives out a yellow showery sparks and produces a black solid. iron reacts with dilute hydrocloric acid to give iron chloride and hydrogen gas.
Molar mass of NH_3



We know.
No of moles=Given mass/Molar mass


Now
Lets write the balanced equation

- There is 2moles of Ammonia
- 3moles of H_2
- 1mole of N_2
Now

For Hydrogen



For Ammonia



For Nitrogen


Explanation:
Protons have a positive charge. Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.
Answer:
The chiefs
Explanation:
hope my answers is helpful
Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= 
= rate constant at
= 
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)

Therefore, the temperature will be, 392.462 K