<span>Answer is: the mass of hydrogen is 22,05 grams.
m(</span>Al(C₂H₃O₂)₃)<span> = 500 g.
M</span>(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.<span>
n</span>(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
Answer: 30.06% is your answer.
I hope this helps.
Stay safe and have a good day :D
Arsenic, I believe. Metalloids fall in between metals and nonmetals (usually on the bold line separating the two on the periodic table). And since the metalloid in question has four electron shells and five valence electrons in the outermost shell, you can see that this element is arsenic