[H⁺]=6.696 x 10⁻⁵
pH = 4.174
<h3>
Further explanation
</h3>
Given
The concentration of 0.000295 M (2.95 x 10⁻⁴ M) butanoic acid solution
Required
the [H+] and pH
Solution
Butanoic acid is the carboxylic acid group. Carboxylic acids are weak acids
For weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
[H⁺]=√1.52 x 10⁻⁵ x 2.95 x 10⁻⁴
[H⁺]=6.696 x 10⁻⁵
pH = - log [H⁺]
pH = - log 6.696 x 10⁻⁵
pH = 5 - log 6.696
pH = 4.174
<u>Answer:</u> The given sample of water is not safe for drinking.
<u>Explanation:</u>
We are given:
Concentration of fluorine in water recommended = 4.00 ppm
ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.
To calculate the ppm of fluorine in water, we use the equation:
![\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6](https://tex.z-dn.net/?f=%5Ctext%7Bppm%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%7D%7B%5Ctext%7BMass%20of%20solution%7D%7D%5Ctimes%2010%5E6)
Both the masses are in grams.
We are given:
Mass of fluorine =
(Conversion factor: 1 g = 1000 mg)
Mass of water = 5.00 g
Putting values in above equation, we get:
![\text{ppm of fluorine in water}=\frac{0.152\times 10^{-3}}{5}\times 10^6\\\\\text{ppm of fluorine in water}=30.4](https://tex.z-dn.net/?f=%5Ctext%7Bppm%20of%20fluorine%20in%20water%7D%3D%5Cfrac%7B0.152%5Ctimes%2010%5E%7B-3%7D%7D%7B5%7D%5Ctimes%2010%5E6%5C%5C%5C%5C%5Ctext%7Bppm%20of%20fluorine%20in%20water%7D%3D30.4)
As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.
Hence, the given sample of water is not safe for drinking.
Explanation:
We assume that
is represented by A and
is represented by B respectively.
According to Wilke Chang equation as follows.
![D_{AB} = \frac{7.4 \times 10^{-8} \times (\phi_{B} M_{B})^{1/2} \times T}{V^{0.6}_{A} \times \mu_{B}}](https://tex.z-dn.net/?f=D_%7BAB%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BB%7D%20M_%7BB%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BA%7D%20%5Ctimes%20%5Cmu_%7BB%7D%7D)
![D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}](https://tex.z-dn.net/?f=D_%7BO_%7B2%7D%20-%20H_%7B2%7DO%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BH_%7B2%7DO%7D%20M_%7BH_%7B2%7DO%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%20%5Cmu_%7BH_%7B2%7DO%7D%7D)
where, T = absolute temperature = (273 + 37)K = 310 K
= an association parameter for solvent water = 2.26
= Molecular weight of water = 18 g/mol
= viscosity of water (in centipoise) = 0.62 centipoise
= the molar volume of oxygen = 25.6 ![cm^{3}/g mol](https://tex.z-dn.net/?f=cm%5E%7B3%7D%2Fg%20mol)
Hence, putting the given values into the above formula as follows.
![D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}](https://tex.z-dn.net/?f=D_%7BO_%7B2%7D%20-%20H_%7B2%7DO%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BH_%7B2%7DO%7D%20M_%7BH_%7B2%7DO%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%20%5Cmu_%7BH_%7B2%7DO%7D%7D)
= ![\frac{7.4 \times 10^{-8} \times (2.26 \times 18)^{1/2} \times 310 K}{(25.6)^{0.6}_{O_{2}} \times 0.692}](https://tex.z-dn.net/?f=%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%282.26%20%5Ctimes%2018%29%5E%7B1%2F2%7D%20%5Ctimes%20310%20K%7D%7B%2825.6%29%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%200.692%7D)
= ![3021.7 \times 10^{-8} cm^{2}/s](https://tex.z-dn.net/?f=3021.7%20%5Ctimes%2010%5E%7B-8%7D%20cm%5E%7B2%7D%2Fs)
Thus, we can conclude that the diffusion of
in water by the Wilke-Chang correlation at
.
Answer:
it is like when rubbing a balloon on your head the energy from your body and from you moving the balloon fast makes energy to make your hair stand up.
Explanation:
Hope this helps :)
Pass the charged particles at high velocity through an electric field.
The positively charged particles will deflect one way and the negatively charged particles will deflect the opposite way. The deflection will observe the Right-Hand Rule. The rate of detection will also depend on the size of the charged particle with heavier one being deflected more.