Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
Answer:
Certain types of atoms are "radioactive," meaning that they will eventually decay, or "break down" into a different type of atom. In this activity, you will simulate radioactive decay by flipping coins. Coins that land tails-up "decay," and coins that land heads-up remain the same.
Explanation:
#Suoka64
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:
they are single cells that lack a nucleus and tend to live in extreme conditions.- B.
