Answer:
sorry to waste ur time but im getting points bc i have a question and i need more points to say ig
Answer:
<h2>Hereis the correct answer </h2>
(A.-9)
Explanation:
<h3>STUDY CORRECTION. </h3>
Answer:
<h2>0.4 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
![n = \frac{N}{L} \\](https://tex.z-dn.net/?f=n%20%3D%20%20%5Cfrac%7BN%7D%7BL%7D%20%5C%5C)
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
We have
![n = \frac{2.4 {10}^{23} }{6.02 \times {10}^{23} } = \frac{2.4}{6.02} \\ = 0.3986...](https://tex.z-dn.net/?f=n%20%3D%20%20%5Cfrac%7B2.4%20%7B10%7D%5E%7B23%7D%20%7D%7B6.02%20%5Ctimes%20%20%7B10%7D%5E%7B23%7D%20%7D%20%20%3D%20%20%5Cfrac%7B2.4%7D%7B6.02%7D%20%20%5C%5C%20%20%3D%200.3986...)
We have the final answer as
<h3>0.4 moles</h3>
Hope this helps you
Answer:
added water = 171 ml
Explanation:
Assuming volumes are additive, the rule that we will use to solve this question is:
M1V1 = M2V2
where:
M1 is the initial concentration = 0.4 m
V1 is the initial volume = 57 ml
M2 is the final concentration = 0.1 m
V2 is the final volume that we want to calculate
Substitute with the given in the above equation to get V2 as follows:
M1V1 = M2V2
(0.4)(57) = (0.1)V2
22.8 = 0.1V2
V2 = 228 ml
Now, the final volume is equal to the initial volume plus the amount of added water. So, to get the amount of added water, we will subtract the initial volume from the final volume as follows:
V2 = V1 + added water
228 = 57 + added water
added water = 228 - 57 = 171 ml
Hope this helps :)
Answer:
![\large \boxed{\text{2920 yr}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B2920%20yr%7D%7D)
Explanation:
Two important formulas in radioactive decay are
![(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt](https://tex.z-dn.net/?f=%281%29%20%5Cqquad%20t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cdfrac%7B%5Cln%202%7D%7Bk%7D%5C%5C%5C%5C%282%29%20%5Cqquad%20%5Cln%20%5Cleft%28%5Cdfrac%7BN_%7B0%7D%7D%7BN%7D%5Cright%29%20%3D%20kt)
1. Calculate the decay constant k
![\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{1530 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{1530 yr}}\\\\& = & 4.530 \times 10^{-4} \text{ yr}^{-1}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7Dt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%26%3D%26%20%5Cdfrac%7B%5Cln%202%7D%7Bk%7D%5C%5C%5C%5C%5Ctext%7B1530%20yr%7D%20%26%3D%20%26%5Cdfrac%7B%5Cln%202%7D%7Bk%7D%5C%5C%5C%5Ck%20%26%20%3D%20%26%20%5Cdfrac%7B%5Cln%202%7D%7B%5Ctext%7B1530%20yr%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%204.530%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20yr%7D%5E%7B-1%7D%5C%5C%5Cend%7Barray%7D)
2. Calculate the time
![\begin{array}{rcl}\ln \left(\dfrac{N_{0}}{N}\right) &= &kt \\\\\ln \left(\dfrac{15}{4}\right) &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\\\\ln 3.75 &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\t &= &\dfrac{\ln 3.75}{4.530 \times 10^{-4} \text{ yr}^{-1}}\\\\& = & \textbf{2920 yr}\\\end{array}\\\text{It would take $\large \boxed{\textbf{2920 yr}}$ for the rate to decrease to 4 dis/min.}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cln%20%5Cleft%28%5Cdfrac%7BN_%7B0%7D%7D%7BN%7D%5Cright%29%20%26%3D%20%26kt%20%5C%5C%5C%5C%5Cln%20%5Cleft%28%5Cdfrac%7B15%7D%7B4%7D%5Cright%29%20%26%3D%20%264.530%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20yr%7D%5E%7B-1%7D%5Ctimes%20t%20%5C%5C%5C%5C%5Cln%203.75%20%26%3D%20%264.530%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20yr%7D%5E%7B-1%7D%5Ctimes%20t%20%5C%5Ct%20%26%3D%20%26%5Cdfrac%7B%5Cln%203.75%7D%7B4.530%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20yr%7D%5E%7B-1%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B2920%20yr%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20would%20take%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B2920%20yr%7D%7D%24%20for%20the%20rate%20to%20decrease%20to%204%20dis%2Fmin.%7D)