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2 years ago

How does vegetation slow and prevent sediment loss?

1 answer:

5 0

Vegetation cover is the most effective and practical way to prevent loss of sediment. The roots of vegetation such as grass bind soil particles together to resist erosion especially from runoff water. Vegetation helps absorb the impact of raindrops to prevent detachment of soil particles. Vegetation can flatten act like shingles on a roof, which allows the runoff to flow above the soil. Stiff standing vegetation act as a wind break, reducing wind velocity, such that it is not powerful enough to dislodge sediments from the ground surface.

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Explain how manipulation of light waves can cause reflection, refraction, diffusion, and absorption; and describe how different

Effectus [21]

Answer:is A

Explanation:

6 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with

andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

$Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g$

Molar mass of nitrogen gas, $N_2$ = 28.014 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass}{28.014\ g/mol}$

Mass of nitrogen gas = 28.014x g

So,

Let, $Mass_{Container}=y$

$y+28.014x=37.289$

Similarly,

$Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g$

Molar mass of nitrogen gas, $CO_2$ = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass{44.01\ g/mol}$

Mass of nitrogen gas = 44.01x g

So,

$y+44.01x=37.440$

Solving the two equations, we get :

$Mass_{Container}=y=37.025\ g$

x = 0.00943 moles

Thus, Given:

$Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g$

$37.025\ g+Mass_{Unknown\ gas}=37.062\ g$

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00943\ moles= \frac{0.037\ g}Molar mass}$

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0

2 years ago

Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na

Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

$\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

valence according to the amount of salt anion

Input the value :

$\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3$

7 0

2 years ago

50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.

photoshop1234 [79]

Limiting reactant : O₂

Mass of N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

$\tt =\dfrac{50}{44}=1.136$

mol O₂ :

$\tt =\dfrac{50}{32}=1.5625$

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136 1.5625

1.0416 1.5625 1.0416

0.0944 0 1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

$\tt =mol\times MW=1.0416\times 92=95.83~g$

7 0

2 years ago