<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
<span>
</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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</span>
Abhishek Singh present invention to provide
4.5x10^22.
How many molecules are in 2.10 mol CO2? 1.26x10^24 molecules.
What is the molar mass of AuCl3?
Answer:
True.
Explanation:
Btw ya spelled mitosis wrong...
