Answer:
![K = \frac {[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^{2} \cdot [H_{3}O^{+}]^{2}}](https://tex.z-dn.net/?f=%20K%20%3D%20%5Cfrac%20%7B%5BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%5D%7D%7B%5BCrO_%7B4%7D%5E%7B2-%7D%5D%5E%7B2%7D%20%5Ccdot%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%5E%7B2%7D%7D%20)
Explanation:
The equilibrium constant for a given reversible aqueous reaction is defined by the product ratio of the concentrations between the products and reactants:
aA + bB ⇄ cC + dD
![K = \frac {[products]^{p}}{[reactants]^{r}} = \frac {[C]^{c} \cdot [D]^{d}}{[A]^{a} \cdot [B]^{b}}](https://tex.z-dn.net/?f=%20K%20%3D%20%5Cfrac%20%7B%5Bproducts%5D%5E%7Bp%7D%7D%7B%5Breactants%5D%5E%7Br%7D%7D%20%3D%20%5Cfrac%20%7B%5BC%5D%5E%7Bc%7D%20%5Ccdot%20%5BD%5D%5E%7Bd%7D%7D%7B%5BA%5D%5E%7Ba%7D%20%5Ccdot%20%5BB%5D%5E%7Bb%7D%7D%20)
<em>where K: is the equilibrium constant, [C] and [D]: are the product concentrations, [A] and [B]: are the reactant concentrations and a,b,c,d: are the stoichiometric coefficients from the reaction. </em>
Therefore, based on the definition the equilibrium constant of our reaction is:
2CrO₄²⁻(aq) + 2H₃O⁺(aq) ⇄ Cr₂O₇²⁻(aq) + 3H₂O(l)
Generally, the water concentration is omitted from the expressions.
I hope it helps you!
Answer:
4g/mol
Explanation:
Firstly, we can get the number of moles of the gas present using the ideal gas equation.
PV = nRT
Here:
P = 886 torr
V = 224ml = 224/1000 = 0.224L
T = 55 degrees celcius= 55+ 273.15 = 328.15K
R = molar gas constant = 62.36 L⋅Torr⋅K−1⋅mol−1
n = PV/RT
n = (886 * 0.224)/(62.36 * 328.15)
n = 0.009698469964 mole
Now to get the molar mass, this is mathematically equal to the mass divided by the number of moles. We have the mass and the number of moles, remaining only the molar mass.
First, we convert the mass to g and that is 38.8/1000 = 0.0388
The molar mass is thus 0.0388/0.009698469964 = 4g/mol
<u>Answer:</u> The partial pressure of carbon dioxide at equilibrium is 0.0056 atm
<u>Explanation:</u>
The given chemical equation follows:
<u>Initial:</u> 4.00
<u>At eqllm:</u> 4.00-2x x x
The expression of 
for above reaction follows:
The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure
Partial pressure of 
= 0.0056 atm
Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm
Answer: C-Fat deposits put a strain on the body.
