The solubility equilibrium of 
:
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
Nickel is a pure substance
Rust is also a pure substance
Ethics and Skepticism are integral to science because they are the basis for the need to lay explanations on various fields of science through experiment. This will enable science bring the right data and arguments through scientific methods in an accurate and a straight forward manner.
Hope this helped have an awesome day :)
Answer:
Kc = [CH₄] / [H₂]²
Kp = [CH₄] / [H₂]² * (0.082*T)^-1
Explanation:
Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.
<em>Pure solids and liquids are not taken into account in an equilibrium</em>
Thus, for the reaction:
C(s)+ 2H₂(g) ⇌ CH₄(g)
Equilibrium constant is:
<h3>Kc = [CH₄] / [H₂]²</h3>
Now, using the formula:
Kp = Kc* (RT)^Δn
<em>Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)</em>
Replacing:
<h3>Kp = [CH₄] / [H₂]² * (0.082*T)^-1</h3>
<em />
