If a uranium atom undergoes both alpha and gamma decay, then it means that there will be formation of one helium particle which is also known as alpha particle and gamma decay is the radiation or release of energy. Whereas in a radioactive reaction, Uranium-235 absorbs a neutron and splits into two new atoms.
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Answer:
The atomic radius <em>Decreases</em> as you move from left to right on the periodic table.
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<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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Se - 78
Selenium - 78
or
78/34 Se2-
If I understood question correctly,
football 1.3 • 102=132.6
Answer is 132 students participate in football.