The answer is 8 because if you add all of the numbers up you get 40, then you divide by the amount of numbers (which is 5) so 40/5=8
Answer:
I think No.
Explanation:
Because mostly fossils are found in sedimentary rocks.
Relative humidity<span> (</span>RH<span>) is the ratio of the </span>partial pressure<span> of water vapor to the </span>equilibrium vapor pressure<span> of water at a given temperature. Relative humidity depends on temperature and the pressure of the system of interest. It requires less water vapor to attain high relative humidity at low temperatures; more water vapor is required to attain high relative humidity in warm or hot air.</span>
Answer:
A. ![C_2=1.5\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D1.5%5Cfrac%7Bmg%7D%7BmL%7D)
B. ![C_2=0.075\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D0.075%5Cfrac%7Bmg%7D%7BmL%7D)
C. ![C_2=0.01\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D0.01%5Cfrac%7Bmg%7D%7BmL%7D)
D. ![C_2=0.001\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D0.001%5Cfrac%7Bmg%7D%7BmL%7D)
Explanation:
Hello.
In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:
![C_2=\frac{C_1V_1}{V_2}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7BC_1V_1%7D%7BV_2%7D)
Thus, we proceed as follows:
A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.
![C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7B300%5Cmu%20L%2A5%5Cfrac%7Bmg%7D%7BmL%7D%20%7D%7B%28300%2B700%29%5Cmu%20L%7D%5C%5C%5C%5CC_2%3D1.5%5Cfrac%7Bmg%7D%7BmL%7D)
B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.
![C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7B50%5Cmu%20L%2A1.5%5Cfrac%7Bmg%7D%7BmL%7D%20%7D%7B%2850%2B450%2B500%29%5Cmu%20L%7D%5C%5C%5C%5CC_2%3D0.075%5Cfrac%7Bmg%7D%7BmL%7D)
C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.
![C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7B10%5Cmu%20L%2A1%5Cfrac%7Bmg%7D%7BmL%7D%20%7D%7B%2810%2B990%29%5Cmu%20L%7D%5C%5C%5C%5CC_2%3D0.01%5Cfrac%7Bmg%7D%7BmL%7D)
D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.
![C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7B10%5Cmu%20L%2A0.1%5Cfrac%7Bmg%7D%7BmL%7D%20%7D%7B%2810%2B990%29%5Cmu%20L%7D%5C%5C%5C%5CC_2%3D0.001%5Cfrac%7Bmg%7D%7BmL%7D)
Best regards.