Given that the maximum acceptable concentration of fluoride in tap water = 1.5 mg/L.
1 mg/L is equivalent to 1 parts per million(ppm).
Converting 1.5 mg/L to ppm:
So the maximum acceptable concentration of fluoride in tap water in ppm is 1.5 ppm.
Finding out the volume of tap water that would contain 1.0 g fluoride:
Converting 1.0 g fluoride to mg:
Taking the concentration of fluoride in tap water to be 1.5 mg/L,
Volume of tap water that contains 1000 mg fluoride
=
Rounding the volume to three significant figures, 667 mL of tap water.
The correct answer is isobar
Explanation:
It is known that the specific heat capacity of Liver 
is 3.59 kJ 
It is given that :
Initial temperature of Liver = Body temperature = 
= 310 K
Final temperature of Liver = 180 K
Relation between heat energy, mass, and change in temperature is as follows.
Q = 
Now, putting the given values into the above formula as follows.
Q =
Q = 
= 700.05 kJ
Therefore, we can conclude that amount of heat which must be removed from the liver is 700.05 kJ.
RMM of magnesium oxide=24+16=40
40g=1mol
therefore 62.5g will have =62.5/40=1.5625 moles
1.5 litres contain 1.5625 mols
therefore 1 litre =1*1.5625/1.5=1.042M
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
