<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]
<span>
</span>
There are 0.31 mol in h3po4
Scientists observe the world around them, from which to draw questions. Their predictions as to the answer are what we call a “hypothesis”. Thus, a scientist’s job is to answer the very hypotheses that they and their piers come up with.
Answer:
9
Explanation:
To balance the equation, we need to supply the number of missing coefficients.
There are 2 moles of aluminium on the right hand side, so automatically we need to multiply the number of moles of aluminium by 2 on the left hand side.
There are 3 sulphate ions on the right hand side so we need to multiply the number on the left by 3.
And lastly since we have 6 hydrogen molecules on the left hand side now, we need to multiply the hydrogen on the right hand side by 3 to give 6
The set of coefficients = 2 3 1 3
Adding all thus yields 9
Answer:
27.9 g
Explanation:
CsF + XeF₆ → CsXeF₇
First we <u>convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles</u>, using its<em> molar mass</em>:
- Molar mass of CsXeF₇ = 397.193 g/mol
- 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇
As <em>1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇</em>, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.
Now we <u>convert 0.184 moles of CsF to moles</u>, using the <em>molar mass of CsF</em>:
- Molar mass of CsF = 151.9 g/mol
- 0.184 mol * 151.9 g/mol = 27.9 g
