Answer:
1.75M
Explanation:
The problem deals with finding the molarity of the given compound.
The compound is:
Na₂SO₄.10H₂O
Mass of the compound = 10.5kg = 10500g
Volume of the compound = 18.6L
Molarity is the number of moles of solute in a solution;
Molarity = 
Number of moles = 
Molar mass of Na₂SO₄.10H₂O = 2(23) + 32 + 4(16) + 10[2(1) + 16]
= 322g/mol
Now;
Number of moles = 
= 32.6mole
So;
Molarity = 
= 1.75M
U need positive and negative electordoe and an electrolyte <span />
Hey there!
27 tells us the sum of protons and neutrons is 27.
Al tells us we have 13 protons.
3- tells us that there are 3 more electrons than protons.
13 + n = 27
neutrons = 14
13 + 3 = 16 electrons
27Al3- has 13 protons, 14 neutrons, and 16 electrons.
Hope this helps!
Answer:
b. 
Explanation:
Hello there!
In this case, given the ionization reaction of HClO as weak acid:
We can write the equilibrium expression as shown below:
![Ka=3.5x10^{-8}=\frac{[H^+][ClO^-]}{[HClO]}](https://tex.z-dn.net/?f=Ka%3D3.5x10%5E%7B-8%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5BClO%5E-%5D%7D%7B%5BHClO%5D%7D)
In such a way, via the definition of x as the reaction extent, we can write:
![3.5x10^{-8}=\frac{x^2}{[HClO]}](https://tex.z-dn.net/?f=3.5x10%5E%7B-8%7D%3D%5Cfrac%7Bx%5E2%7D%7B%5BHClO%5D%7D)
As long as Ka<<<<1 so that the x on the bottom can be neglected. Thus, we solve for x as shown below:
And finally the percent dissociation:
![\% diss=\frac{x}{[HClO]} *100\%\\\\\% diss=\frac{7.94x10^{-5}M}{0.18}*100\% \\\\\% diss =0.044\%=4.4x10^{-2}\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHClO%5D%7D%20%2A100%5C%25%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B7.94x10%5E%7B-5%7DM%7D%7B0.18%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D0.044%5C%25%3D4.4x10%5E%7B-2%7D%5C%25)
Which is choice b.
Best regards!
