T½=18.72days
therefore t¾=18.72+½ of 18.72
we have 18.72+9.36=28.08days
First factor: Density of water
This density of water is affected by the level of salt present in it. They are inversely proportional. This means that as the amount of salt in water increases (water becomes more salty), the density of water would decrease
Second factor: Temperature
Temperature affects the composition of water because as temperature increases, the ocean water evaporates and the salt precipitates. Therefore, this factor is essential in determining the rate by which the ocean water evaporates
Third factor: Salinity
This factor helps in determining the amount of salt present in the ocean, i.e, concentration of salt in the water.
Hope this helps :)
Answer:
red blood cell → heart tissue → circulatory system
Explanation:
The levels of organization from the simplest level of organization to the most complex is cell, tissue, organ, organ, organ system, and organism. Red blood cells are cells, heart tissue is tissue in the heart and the circulatory system is an organ system that ensures blood, oxygen, and nutrients are flowing through the body.
This problem is about conversion and dimensional analysis. Important information to know:
1 atm = 760 torr = 101.325 kPa
For atm to torr conversion:
0.875 atm * (760 torr / 1 atm) = 665 torr
For atm to kPa conversion:
0.875 atm * (101.325 kPa / atm) = 88.7 kPa
Thus the answer is b) 665 torr and 88.7 kPa
Answer: The correct option is, (C) 0.53
Explanation:
The given chemical reaction is:
The rate of the reaction for disappearance of A and formation of C is given as:
![\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DA%3D-%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
Or,
![\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DC%3D%2B%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where,
= change in concentration of C = 1.33 M
= change in time = 4.5 min
Putting values in above equation, we get:
![\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B1.33M%7D%7B4.5min%7D)
![\frac{\Delta [A]}{\Delta t}=0.53M/min](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D0.53M%2Fmin)
Thus, the decrease in A during this time interval is, 0.53
