Answer:
1. 7 (a neutral solution)
Answer: 10-7= 0.0000001 moles per liter
2. 5.6 (unpolluted rainwater)
Answer: 10-5.6 = 0.0000025 moles per liter
3. 3.7 (first acid rain sample in North America)
Answer: 10-3.7 = 0.00020 moles per liter
The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.
Explanation:
The molar concentration of a solution, usually expressed as the number of moles of solute per liter of solution.
Answer:
the particles of matter are arranged in different ways for the different states
Explanation:
because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration
Answer:
The proportion of dissolved substances in seawater is usually expressed in ppm, ppb or ppt
Explanation:
The concentration of very diluted solutions should be expressed in parts per million, billion or trillion.
ppm = mass from the solute . 10⁶ / mass or volume of the solution
ppb = mass from the solute . 10⁹ / mass or volume of the solution
ppt = mass from the solute . 10¹² / mass or volume of the solution
ppm = mg/kg, μg/g, μg/mL → These are the units
ppb = ng/g
ppt = pg/g
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)
![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n
n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
