<em>K</em>_eq = 0.14
The chemical equation is
A ⇌ B
The equilibrium constant expression is
<em>K</em>_eq = [B]/[A]
If [A] = 7[B]
<em>K</em>_eq = [B]/{7[B]}= 1/7 = 0.14
Answer:
The precipitate will form.
Explanation:
Let's write the equilibrium expression for the solubility product of calcium sulfate:
⇄ 
The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:
![K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D10%5E%7B-4.5%7D)
Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:
![Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}](https://tex.z-dn.net/?f=Q_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D5.00%5Ccdot10%5E%7B-2%7D%20M%5Ccdot7.00%5Ccdot10%5E%7B-3%7D%20M%3D3.5%5Ccdot10%5E%7B-4%7D)
Now given the K value of:

Notice that:

This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.
Answer:
32.9%
Explanation:
<u>The mass of water </u>present in the original sample can be calculated by the substraction:
- m H₂O = sample mass before heating - sample mass after heating
- m H₂O = 8.24 g - 6.20 g = 2.04 g
The <u>percent by mass of water</u> can be calculated with the formula:
- Water Mass / Original Sample Mass * 100 %
* 100% = 32.9%
The atomic number is the number of protons