Answer:
I think it's 5. But I'm not sure what you mean by right-hand side...
Explanation:
I really hope this helps you!
Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O) = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.
It is softer than topaz and and it is softer than diamond (diamond has a Mohs hardness of 10, which is the highest value of the scale)
<u>Answer:</u> The expression for equilibrium constant in terms of concentration is ![K_c=[CO_2]](https://tex.z-dn.net/?f=K_c%3D%5BCO_2%5D)
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by 
For a general chemical reaction:

The
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1.
For the given chemical reaction:

The expression for
is:
![K_{c}=\frac{[MgO][CO_2]}{[MgCO_3]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BMgO%5D%5BCO_2%5D%7D%7B%5BMgCO_3%5D%7D)
In the above expression, magnesium oxide and magnesium carbonate will not appear because they are present in solid state.
So, the expression for
becomes:
![K_{c}=[CO_2]](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5BCO_2%5D)
Hence, the equilibrium constant for the reaction is given above.