Answer:
0.5188 M or 0.5188 mol/L
Explanation:
Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.
***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".
The formula for molarity is:
n = moles (unit mol)
V = volume (unit L)
<u>Find the molar mass (M) of potassium hydroxide.</u>
![M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}](https://tex.z-dn.net/?f=M_%7BKOH%7D%20%3D%20%5Cfrac%7B39.098%20g%7D%7Bmol%7D%2B%5Cfrac%7B16.000%20g%7D%7Bmol%7D%2B%5Cfrac%7B1.008%20g%7D%7Bmol%7D)
<u>Calculate the moles of potassium hydroxide.</u>
![n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}](https://tex.z-dn.net/?f=n_%7BKOH%7D%20%3D%20%5Cfrac%7B14.555%20g%7D%7B1%7D%2A%5Cfrac%7B1mol%7D%7B56.106g%7D)
![n_{KOH} = 0.25941(9)mol](https://tex.z-dn.net/?f=n_%7BKOH%7D%20%3D%200.25941%289%29mol)
Carry one insignificant figure (shown in brackets).
<u>Convert the volume of water to litres.</u>
![V = \frac{500.0mL}{1}*\frac{1L}{1000mL}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B500.0mL%7D%7B1%7D%2A%5Cfrac%7B1L%7D%7B1000mL%7D)
![V = 0.5000L](https://tex.z-dn.net/?f=V%20%3D%200.5000L)
Here, carrying an insignificant figure doesn't change the value.
<u>Calculate the concentration.</u>
![c = \frac{n}{V}](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7Bn%7D%7BV%7D)
<= Keep an insignificant figure for rounding
<= Rounded up
<= You use the unit "M" instead of "mol/L"
The concentration of this standard solution is 0.5188 M.
"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered
A Bronsted-Lowry acid is a chemical species that donates one or more hydrogen ions in a reaction. In contrast, a Bronsted-Lowry base accepts hydrogen ions. When it donates its proton, the acid becomes its conjugate base. A more general look at the theory is as an acid as a proton donor and a base as a proton acceptor. :)
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
the answer is C. the solute polar water itself is a polar solvent so it combines well with other polar solvents (which basically have polar molecules)