Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
Now, find moles of 
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
I would believe it to be 55g. A -> B YIELDS AB. So, 10g + 45g = 55g.
Answer:
a )Li
b)O
c)F
Explanation:
a) Li-1s^2 2s^1
F-1s^2 2s^2 2p^5
it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy
b)oxygen ion is larger than Na because o have fewer proton
c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.
Moles = n/v where n is the moles of solute and v being the liters of solution.
We can put in the information provided to find the molarity.
Moles = .45/3.0 = .15
So we now know that the molarity of that solution is .15!
I hope I helped you :). Make sure to memorize that formula because it's not that hard as long as you know what to plug in.
