Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Molarity of the resulting solution will be 1.33 M.
<u>Explanation:</u>
First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume
For cup 1 = 1 M ×0.05 L = 0.05 moles
For cup 2 = 2.5 M × 0.05 L= 0.125 moles
For cup 3 = 0.5 M × 0.05 L = 0.025 moles
Total moles = 0.05 + 0.125 + 0.025 = 0.2 moles
We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L
Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M
0 because you would be floating at the top within seconds and also have fun with the sudden stop
I’m confused what are you asking exactly?
Answer:
CaO
Explanation:
Firstly, let’s get the percentage compositions.
The total mass is now = 4.008 + 1.600 = 5.608g
The percentage composition of carbon is thus 4.008/5.608 * 100 = 71.47%
The percentage composition of oxygen is 100 - 71.4 = 28.53%
Now, we divide each percentage by the atomic masses. The atomic mass of calcium is 40 while that of oxygen is 16
Ca = 71.47/40 = 1.78675
O = 28.53/16 = 1.783125
Both values are quite similar and dividing by the smallest will yield same values of 1.
Hence , the empirical formula is caO
