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2 years ago

How does the vegetation surface type affect the amount of runoff

Chemistry

1 answer:

maxonik [38]2 years ago

5 0

Answer:

The type of vegetation a surface does affect the water coming from above to sink in or runoff. 

Explanation:

This is how the vegetation affects the runoff:-

The leaves and stems present in the vegetation do not let the water fall directly on the soil and makes the process rather slow which makes the water to get to the ground slowly and sink in properly inside the soil rather than running off.

If the vegetation present is dense with there was being hairy then also the water would not run out and will get absorbed by the roots letting the soil intact

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Which of the following gases experience the greatest degree of deviation from ideal behavior under high pressures and low temper

xz_007 [3.2K]

A) CH4

In general, methane reactions are difficult to control. Partial oxidation to methanol, for example, is a rather difficult reaction because the chemical reactions that occur continue to form carbon dioxide and water even though the amount of oxygen available is insufficient.

<h2>Further explanation </h2>

Methane is the simplest hydrocarbon in the form of gas with the chemical formula CH4. Pure methane does not smell, but if used for commercial purposes, a bit of sulfur is usually added to detect leaks that might occur.

Methane is a greenhouse gas. Methane is used in chemical industrial processes and can be transported as frozen liquids (liquefied natural gas, or LNG).

Methane is a major component of natural gas, around 87% of volume.

Methane is not toxic, but is highly flammable and can cause explosions when mixed with air.

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Details

Class: college

Subject: chemistry

Keywords: ch4, methane, chemicals

8 0

2 years ago

A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

3 years ago

Octopus and squids breathe through

svet-max [94.6K]

Octopus and squids breathe like fishes they breathe from gills 

so even octopus and squids breathe through gills too.

maybe this answer would help u

4 0

2 years ago

Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.

Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

$C(gr) +H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The combustion of liquid acetic acid is given as,

$CH_{3} COOH(l) + 2O(g)$ → $2CO_{2}((g) +2H_{2} O(l)$ ΔH =-873

learn more about balancing chemical equation

brainly.com/question/15052184

#SPJ4

5 0

1 year ago

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

$\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}$

3 0

2 years ago