Answer:
You get some type of pressure that you start to feel in your muscles and joints from gravity and movement. How do they say it? Something called "seat-of-the-pants" (something like that). You get some type of pressure, and your body senses it, and it knows when you are upside-down or not, because if you're not, then you won't get any pressure in your muscle.
Hope this helped!
Have a supercalifragilisticexpialidocious day!
Number of Protons in an Uncharged Atom
The two main components of an atom are the nucleus and the cloud of electrons. The nucleus contains positively charged and neutral subatomic particles, whereas the cloud of electrons contains tiny negatively charged particles.
Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Answer:
the answer should be henry's law
Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2
The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
