Answer:
the chemical symbol is 96
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :

a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
Answer:

Explanation:
Hello,
In this case, the law of mass action for the first reaction turns out:
![Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%3D1.27)
Now, for the second reaction is:
![Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D)
Therefore, by applying square root for the first reaction, one obtains:
![\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Csqrt%7BKc%7D%20%3D%5Csqrt%7B%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%7D%20%3D%5Csqrt%7B1.27%7D)
![\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5BAsH_3%5D%5E2%7D%20%7D%7B%5Csqrt%7B%5BAs%5D%5E2%7D%20%5Csqrt%7B%5BH_2%5D%20%5E3%7D%20%7D%20%3D%5Csqrt%7B1.27%7D)
![\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}](https://tex.z-dn.net/?f=%5Csqrt%7B1.27%7D%3D%5Cfrac%7B%5BAsH_3%5D%7D%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D)
Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:
![Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1.27%7D%20%7D%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D%3D0.887)
Best regards.
Answer:We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
XeF₆
Explanation:
Im not sure how correct these are but i hope they help :)
some potential barriers are
youd have to wait for the chemicals to clear up before goiun gback into that room if u have kids that r little you would have to make sure they dont lick wutever it is u cleaned otherwise it might poison them and if u use to much of it it could make the air poisoness for a few which could be dangerous toi breathe in.
again im not to sure how correct these r but i hope they were right and i hope they help :)