The chemical change that involves breaking down substances using electricity is C. Electrolysis.
<span>1. rate = k[A][B]² - is the best choice,
because when you double A. </span>rate = k[2A][B]²=rate = k*2*[A][B]²
when you double B . rate = k[A][2B]²=rate = k[A][B]²*2²= rate = k[A][B]²*4
<span>2.
1) rate1 =[0.20]^x*[0.15]^y=2.4*10⁻²
rate2=[0.20]^x*[0.30]^y=4.8*10⁻²
divide equation of rate2 by rate 1
rate2/rate1 </span>[0.20]^x*[0.30]^y/([0.20]^x*[0.15]^y)=4.8*10⁻²/2.4*10⁻²
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</span> [0.30]^y/*[0.15]^y=4.8/2.4 , ( [0.30]/[0.15])^y=2, 2^y=1 y =1 , so exponent for [Y] will be 1
2)rate1= [<span>0.20]^x*[0.30]^y = 4.8 × 10⁻²
rate2=[0.40]^x *[0.30 ]^y= 19.2 × 10⁻²
</span>divide equation of rate2 by rate 1
rate2/rate1 [0.40]^x*[0.30]^y/([0.20]^x*[0.30]^y)=19.2*10⁻²/4.8*10⁻²
[0.40]^x/[0.20]^x=19.2/4.8
([0.40]/[0.20])^x= 4,
(2)^x=4, x=2, so so exponent for [X] will be 2
3) rate=k[X]²[Y]
4) to find k
take <span> [X]=0.20 M, [Y]= 0.30 M rate=4.8 × 10⁻²</span> M/min
rate=k[X]²[Y]
4.8 × 10⁻² M/min=k[0.20M]²[0.30M]
4.8 × 10⁻² M/min=k*(0.04*0.3)M³
k=(4.8 × 10⁻² M/min)/(0.012 M³)= 4 min/M²
5) final equation
rate=(4 min/M²)*k[X]²[Y]
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The correct option is this: OXYGEN WILL HAVE MORE KINETIC ENERGY THAN NITROGEN.
Increasing the temperature of oxygen requires the application of heat. The heat energy that is applied to the gas will make the particles of the oxygen gas to gain more kinetic energy and to move more rapidly than before, by so doing, the particles will colloid more with one another and with the wall of the container. The kinetic energy of the particles of the nitrogen gas will remain the same since its temperature was not affected.
Answer:
Explanation:
The middle carbon which is in between the two carbonyl groups. One of the carbonyl group is a ketone whereas the other carbonyl group is of an ester.
Since both the carbonyl groups are appreciably electron-withdrawing and hence they withdraw the electron density from the central carbon atom to quite a extent and hence making the hydrogens quite acidic.
Also the carbanion generated on the central carbon atom after the proton abstraction can be appreciably stabilized by conjugation on the two carbonyl groups. Due to this stabilization the carbanion happens to be stable and attacks the b position of of a,b unsaturated carbonyl compound .
Kindly refer the attchment for the reaction and structure.
Michael addition is a reaction of stabilized enolate which reacts to give a conjugate addition.
<span>It showed that a nucleus occupies a small part of the whole atom. The fact that most of the particles passes right through the gold foil indicated that the actual volume of the nucleus is incredibly small compared to the size of the atom since the nucleus is what would stop the particles from passing through it.</span>