830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL
I am guessing that the concentration of your solution is 2.3 mol/L.
a) Moles of CaCl2
MM of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
b) Volume of solution
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).
- (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

- (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

- (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:
%error=
%error=
%error=2.44 %
H = 1 amu
P = 31 amu
O = 16 amu
Therefore:
H3PO4 = 1 x 3 + 31 + 16 x 4 => 98 u
hope this helps!
Answer:
47.3 ml
Explanation:
The graduated cylinder is shown in the image attached.
Now we have to take a good look at the cylinder, the lines between 45 and 50 are 46, 47, 48 and 49. Even though the points in between two lines weren't graduated but we can intelligently guess the correct volume by observing the upper meniscus of the liquid. Hence the answer.
The ________ orbital is degenerate with 5py in a many-electron atom.
<h2>
5px is the correct answer</h2>