I think it might be the last answer.... Or the second one. Yeah i think it’s the second one
<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>
<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M r<span>(t)</span>=0.66</span> M/s
<span>0.2 M r<span>(t)</span>=1.32</span> M/s
<span>0.3 M r<span>(t)</span>=1.98</span> M/s
We can have the relationship:
<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2
Also, we assume m=1 which is the order of the reaction.
Thus, the relationship is simplified to,
r(t)/r0(t) = 2
r<span>(t)</span>=k<span>[<span>N2</span><span>O3</span>]</span>
0.66 <span>M/s=k×0.1 M</span>
<span>k=6.6</span> <span>s<span>−<span>1</span></span></span>
The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required
I think its A.
if one force cannot overcome the other, the object remains stationary.
255 Newtons hope this helps
