Given:
<span>M1 = 6.5 kg of books
</span><span>M2 = 1.5 kg of CDs
</span><span>M3 = 2.0 kg of DVDs
Required: percent by mass of each object
Solution:
First, we calculate the total mass.
M = 6.5 kg + 1.5 kg + 2.0 kg = 10 kg
Percent by mass is calculated by getting the ration of the mass of an object and the total mass multiplied by 100 to get the percent.
%M1 = 6.5 / 10 x 100 = 65%
%M2 = 1.5/10 x 100 = 15%
%M3 = 2.0/10 x 100 = 20%</span>
Answer:
315.51g/mol
Explanation:
137(33 + (16.00 + 1.01) 2 + 8 [1.01 (2) + 16.00] = 315.51g/mol
Answer:
B.
Explanation: hope this helps
Benzene
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