Add exactly 5.00 mL of 0.0400 M stock CuSO4 solution directly into the evaporating dish. Set a 250-mL beaker, Add 100 mL of water. Data for Part I
Mass of empty dish 32.470 g empty dish
Volume of 0.0400 M solution 5.00 mL CuSO4 solution
Mass of dish and 0.0400 M solution 37.497 g dish and solution
Mass of dish and CuSO4 solid 32.503 g dish and CuSO4 solid
Calculations for Part I
1. Calculate the mass of solution
2. Calculate the mass of solid CuSO4 dissolved in the solution.
3. Calculate the number of moles of solid CuSO4 dissolved in the solution.
4. Calculate the mass of water evaporated from the solution.
5. Calculate the density of solution, (g solution/mL solution).
6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution).
7. Calculate the molality of solution (moles CuSO4/kg solvent).
8. Calculate the molarity of solution (moles CuSO4/L solution).
9. Given that the true molarity is 0.0400 M, calculate the percent error of your result.

1) 5.027 g

2) 0.033 g

3) 2.1×10^-4 moles

4) 4.994 g

5) 0.048 g/ml

6) 0.66%

7) 0.042 moles/Kg

8) 2.1 ×10^-3 mol/L

9) 94.75%

Explanation:

1) mass of solution= mass of solution and dish - mass of dish = 37.497 g- 32.470 g = 5.027 g

2) mass of solid dissolved = mass of dish and solid - mass of dish = 32.503 g - 32.470 g= 0.033 g

3) number of moles = mass of CuSO4/Molar mass of CuSO4 = 0.033g/160gmol-1 = 2.1×10^-4 moles

4) mass of water = mass of solution - mass of solute = 5.027g - 0.033 g = 4.994 g

5) density of solution= mass solution/ volume of solution = 4.994 g/100ml = 0.04994 g/ml

6) % by mass of CuSO4= mass of solid / mass of solution × 100 = 0.033g/5.027g × 100 = 0.66%

7) molality of CuSO4= 2.1×10^-4 moles / 5.027×10^-3 = 0.042 moles/Kg

8) molarity = 2.1×10^-4 moles × 1000/100 = 2.1 ×10^-3 mol/L

9) % error = actual - calculated/ actual × 100 = 0.04 - 2.1 ×10^-3 M/0.04 × 100 = 94.75%

answer: h+ its less than oh

explanation:

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."