The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3
Nitrogen (around 78%), Oxygen (around 21%), and Argon (around 1%).
Hope this helps :)
Answer: 1. 
moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 
moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove 
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide= 
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .
So what do you want me to do
Explanation
<h3><u>Answer;</u>
</h3>
= 607.568 Torr
<h3><u>Explanation;
</u></h3>
1 in of mercury is equivalent to 25.4 Torr
Therefore;
23.92 InHg will be equal to;
23.92 × 25.4
<u>= 607.568 Torr</u>
