He proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another.
According to one acid-base theory, a water molecule acts as an acid when the water molecule (3) donates an H+.
Answer:
<h2>15 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
![density = \frac{mass}{volume} \\](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%20%5C%5C%20)
But from the question
volume = final volume of water - initial volume of water
volume = 165 - 150 = 15 mL
We have
![density = \frac{225}{15} = 15 \\](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7B225%7D%7B15%7D%20%20%3D%2015%20%5C%5C%20)
We have the final answer as
<h3>15 g/mL</h3>
Hope this helps you
Answer:
1) 0.18106 M is the molarity of the resulting solution.
2) 0.823 Molar is the molarity of the solution.
Explanation:
1) Volume of stock solution = ![V_1=11.00 mL](https://tex.z-dn.net/?f=V_1%3D11.00%20mL)
Concentration of stock solution = ![M_1=0.823 M](https://tex.z-dn.net/?f=M_1%3D0.823%20M)
Volume of stock solution after dilution = ![V_2=50.00 mL](https://tex.z-dn.net/?f=V_2%3D50.00%20mL)
Concentration of stock solution after dilution = ![M_2=?](https://tex.z-dn.net/?f=M_2%3D%3F)
( dilution )
![M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M](https://tex.z-dn.net/?f=M_2%3D%5Cfrac%7B0.823%20M%5Ctimes%2011.00%20mL%7D%7B50%20%2C00%20mL%7D%3D0.18106%20M)
0.18106 M is the molarity of the resulting solution.
2)
Molarity of the solution is the moles of compound in 1 Liter solutions.
![Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BMass%20of%20compound%7D%7D%7B%5Ctext%7BMolar%20mas%20of%20compound%7D%5Ctimes%20Volume%20%28L%29%7D)
Mass of potassium permanganate = 13.0 g
Molar mass of potassium permangante = 158 g/mol
Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)
![Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B13.0%20g%7D%7B158%20g%2Fmol%5Ctimes%200.100%20L%7D%3D0.823%20mol%2FL)
0.823 Molar is the molarity of the solution.