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3 years ago

Filtration of wastewater flowing over a surface of land with vegetation and water is a

Chemistry

1 answer:

NISA [10]3 years ago

3 0

Filtration of wastewater flowing over a surface of land with vegetation and

water is a biodegradable treatment.

Biodegradable substances are those which can be recycled for other uses

such as manure etc. The filtration of the waste water helps to remove

organic materials and pathogens present in it. The organic materials

removed can then be used for other purposes.

The pathogens are then destroyed to prevent the spread of diseases within

the area. The explanation makes biodegradable treatment most appropriate.

Read more on biodegradable treatment on brainly.com/question/1172912

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help asap What type of cell does the cheek cell represent, plant cell or animal cell? What did you see that let you know?

vladimir1956 [14]

Answer:

The human cheek cell is a good example of a typical animal cell. It has a prominent nucleus and a flexible cell membrane which gives the cell its irregular, soft-looking shape.

7 0

3 years ago

**MEDAL AND FAN** what structural units make up metallic solids?

Margarita [4]

C)

Metallic solids consist of metal ions which are surrounded by delocalized electrons.

8 0

3 years ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

3 0

4 years ago

Calculate the mass of 0.750 mol of the following substance. Na3PO4.

Nookie1986 [14]

Na3 = 23*3 = 69
P = 31
O4 = 16*4 =64
69+ 31+ 64 = 164g/mol
0.75mol = 0.75 * 164 = 123 g

4 0

3 years ago

Swer all parts. (a) What is the wavelength (in nm) of radiation that has an energy ...

bagirrra123 [75]

Answer:

a) The wavelength is around 33.8 nm = 3.38*10⁻¹ nm

b) Ultraviolet

Explanation:

a) The energy (E) of a photon is related to its wavelength (λ) by the Planck's equation:

$E = h\frac{c}{\lambda}----(1)$

where h = Planck's constant = 6.626*10^-34 Js

c = speed of light = 3*10^8 m/s

E = 3.55*10^6 J/mol

The energy in terms of J/photon is:

$=\frac{3.55*10^{6} J/mol*1mol}{6.023*10^{23} photons } =5.89*10^{-18} J/photon$

Based on eq(1)

\lambda = h\frac{c}{E}=6.626*10^{-34}Js*\frac{3*10^{8}m/s}{5.89*10^{-18}J}=3.38*10^{-8}m

The wavelength is around 33.8 nm = 3.38*10⁻¹ nm

b) In the electromagnetic spectrum the ultraviolet range extends from 390 nm-8.82 nm

The calculated wavelength of 33.8 nm should fall in the UV range.

5 0

3 years ago