<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>
<span>False,
This is because when you can easily ionize and atom or the chances of it being ionizable are quite high, it means that that particular atom have very low ionization potential that is the reason why it was easily ionizable
An atom with a high ionization power and a firmly negative electron fondness will both pull in electrons from different particles and oppose having its electrons taken away; it will be an exceedingly electronegative molecule.</span>
Answer: Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.
Hope this helps! :)
Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option
Watermelon And that’s the answer
