Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock = 
molarity of stock = 4M
volume = 
= 0.0625L or 62.5mL
Answer:
Element is made up of the atoms or isotopes. Isotopes are where only neutrons change in number, such as 3 turning to 2 or to 5. Atom is where it is not changed and it is the original, and example is instead of Carbon13 it is just Carbon. I hope this helps.
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:
When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:
1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
I, D, B, E, H, C, J, F, A, G, K
