Α.fℓσυя αи∂ ѕυgαя αяє ѕιfтє∂ тσgєтнєя ιитσ α мєαѕυяιиg ¢υρ
Density = mass/volume = 80 g/10cm³ = 8 g/cm³
Eyepiece, finder-scope, optical tube, aperture, focuser, and mount
Answer:
A. 0.0440 moles/day
Explanation:
First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:
33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag
Now, let's divide this by 7 to get the rate per day:
0.3081 mol Ag ÷ 7 days = 0.0440 mol/day
Thus, the answer is A.
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>
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</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M