<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 answer is B Solubility Increases
Answer:
It can't be done.
Explanation:
If you have only 5.4 g of oxygen, the most lithium oxide you can get is 7.7 g.
Only 2.3 g of lithium will react. and the other 22.3 g of lithium will not be used.
✡ Answer: 1.23*10^2 ✡
- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23
- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2
Final answer: 1.23*10^2
✡Hope this helps✡
