Answer:
3.0 g daily dose
Explanation:
First of all, the given dose is measured in milligrams per kilogram of body and the mass of this patient is given in pounds. This means we have to begin this problem by converting the units.
Convert the mass into kilograms:
![224~lbs\cdot \frac{1~kg}{2.205~lb} = 101.6~kg](https://tex.z-dn.net/?f=224~lbs%5Ccdot%20%5Cfrac%7B1~kg%7D%7B2.205~lb%7D%20%3D%20101.6~kg)
Knowing that the dose is 15.0 mg per kilogram of body, one dose is equal to:
![101.6~kg\cdot \frac{15.0~mg}{1~kg} = 1524~mg](https://tex.z-dn.net/?f=101.6~kg%5Ccdot%20%5Cfrac%7B15.0~mg%7D%7B1~kg%7D%20%3D%201524~mg)
This is a single dose. If we wish to calculate the daily dose, since it's used twice, we get:
![1524~mg/dose\cdot 2~doses/day = 3048~mg = 3.0~g](https://tex.z-dn.net/?f=1524~mg%2Fdose%5Ccdot%202~doses%2Fday%20%3D%203048~mg%20%3D%203.0~g)
Answer:
Any matter considered to be a fuel contains chemical energy
Energy has to be provided to initiate the reaction.
So yes. It is
Answer:
explanation and image attached
Explanation:
Our aim is to draw a structure of XeO2F2 whith the least formal charges. We must remember that the compound has 34 valence electrons.
To obtain the least formal charges then Xe must have a total of twelve electrons on its valence shell instead of eight.
The other atoms around the central Xe atom are arranged as shown in the image attached.
Image Credit: UCLA
Answer:
Here it is (sorry its late)
Explanation:
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Answer:
P[H2] = P[I2] = 0.02 atm
P[HI] = 5.98 atm
Explanation:
The ICE table for the given reaction is:
H2(g) + I2(g) ↔ 2HI(g)
Initial 3.01atm 3.01atm -
Change -x -x +2x
Equilib (3.01-x) (3.01-x) 2x
The equilibrium constant, Kp is given as:
![Kp = \frac{P_{HI}^{2}}{P_{H2}*P_{I2}}](https://tex.z-dn.net/?f=Kp%20%3D%20%5Cfrac%7BP_%7BHI%7D%5E%7B2%7D%7D%7BP_%7BH2%7D%2AP_%7BI2%7D%7D)
![483*10^{2} = \frac{(x)^{2}}{(3.01-x)^{2}}](https://tex.z-dn.net/?f=483%2A10%5E%7B2%7D%20%3D%20%5Cfrac%7B%28x%29%5E%7B2%7D%7D%7B%283.01-x%29%5E%7B2%7D%7D)
x = 2.99 atm
Equilibrium partial pressure of H2 = I2 = 3.01-2.99 = 0.02 atm
Equilibrium partial pressure of HI = 2x = 2(2.99) = 5.98 atm