What would be a simple way to prevent erosion

Answers

Answer 1

Answer: There are many simple ways to prevent erosion. One of the simplest forms of preventing erosion is with vegetation and by establishing root systems that hold the soil together. You can also use geotextiles, fertilizers and also build retainer walls to keep the soil intact and prevent erosion.

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Minnesota is one of the largest contributors of nitrogen to the Gulf. a. True b. False

Mr. Smith wanted to see if the color of light shined on a plant had an effect on the number of leaves it had. He gathered 2 groups of the same species of plants, gave them the same amount of water, and did the test for the same amount of time. On one group of plants he used white light. For the second group he changed the light color to red.

Answers

Final answer:

The color of light affects plants' growth and leaf production due to the role of different light colors in photosynthesis and the phytochrome system in plants. Chlorophyll mainly absorbs red and blue light from the light spectrum for photosynthesis, and far-red light can slow plant growth. Therefore, exposure to different light colors could result in different numbers of leaves.

Explanation:

The subject of the question is how the color of light affects the growth of plants, specifically the number of leaves plants produce. This experiment is primarily about the effect of light on plant's photosynthesis, a process that uses light to convert carbon dioxide and water into glucose. The glucose is utilized to fulfill various energy requirements of the plant's growth, including the growth of leaves.

Light has diverse wavelengths, with each color representing a different wavelength. While white light contains all the possible color wavelengths, the red light specifically contains a higher wavelength region. According to Sir Isaac Newton's experiment, sunlight, which looks white to us, contains all the colors of the spectrum.

In this context, the phytochrome system in plants plays a crucial role. Chlorophyll, the green pigment in leaves, absorbs mainly red and blue light from the light spectrum and uses that energy for photosynthesis. However, far-red light, which is one element of white light, is not absorbed. If a plant is exposed more to far-red light, it could slow its growth.

Consequently, the difference in the number of leaves that Mr.Smith observed in the plants under white light and those under red light might be due to the role of light color in photosynthesis and the phytochrome system in plants.

Learn more about Effect of Light Color on Plant Growth here:

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The main energy products currently being obtained from the ocean floor are oil and _____.

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The main energy products currently being obtained from the ocean floor are oil and natural gas. Both oil and natural gas are resources that are found naturally on Earth. They are both also comprised of hydrocarbons, meaning carbon and hydrogen. The difference lies in their natural physical state. Oil is a liquid, while natural gas is a gas/vapor.

How many moles of NH3 can be produced from 13.5mol of H2 and excess N2??

Answers

Hello there.

How many moles of NH3 can be produced from 13.5mol of H2 and excess N2??

9 moles NH3

What is the theoretical yield of hydrogen gas if 5.00 mol of zinc are added to an excess of hydrochloric acid?

Answers

the theoretical yield of hydrogen gas is 10 g

calculation

Step 1: write the equation for reaction

Zn + 2 HCl → zncl₂ + H₂

Step 2: use the mole ratio to find the moles of H₂

from equation above Zn : H₂ is 1: 1 therefore the moles of H₂ = 5.00 moles

Step 3: find the theoretical yield(mass) for H₂

mass = moles × molar mass

from periodic table the molar mass of H₂= 1×2 = 2 g/mol

mass = 5 moles × 2 g/mol = 10 grams

Answer:

B. 10.1 g

Explanation:

Which is a product of a condensation reaction? 1. O2 2. CO2 3. H2 4. H2O

Answers

The correct answer is H2O, it is a product of a condensation reaction. It is a type of chemical reaction where two molecules combine to form a larger molecule together with a loss of a small molecule. Usually, this small molecule is water.

Answer:

$H_(2)O$ is a product of a condensation reaction.

Explanation:

In a condensation reaction, two molecules are combined to produced a large molecules along with removal of a water molecule.

For an example, let's consider aldol condensation reaction.

In aldol condensation reaction, an enolate anion reacts and combines with an aldehyde or ketone molecule follwed by dehydration. In that dehydration step , $H_(2)O$ gets removed and an unsaturated carbonyl product is formed.

Usually, $O_(2),CO_(2) and H_(2)$ are produced as a by product of a decomposition reaction.

A piece of sodium metal reacts completely with water as follows: 2Na(s) + 2H2O(l) ⟶ 2NaOH(aq) + H2(g) The hydrogen gas generated is collected over water at 25.0°C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25°C = 0.0313 atm.)

Answers

Answer:

$\large \boxed{\text{0.449 g}}$

Explanation:

1. Gather all the information in one place

M_r: 22.99

2Na + 2H₂O ⟶ 2NaOH + H₂

$\, p_{\text{tot}} = \quad \text{1.00 atm}\np_{\text{H2O}} = \text{0.0313 atm}$

T = 25.0 °C

V = 246 mL

2. Moles of H₂

To find the moles of hydrogen, we can use the Ideal Gas Law:

pV = nRT

(a) Calculate the partial pressure of the hydrogen

$p_{\text{tot}} = p_{\text{H2}} + p_{\text{H2O}}\n\text{1.00 atm} =p_{\text{H2}} +\text{0.0313 atm}\np_{\text{H2}} = \text{0.9687 atm}$

(b) Convert the volume to litres

V = 246 mL = 0.246 L

(c) Convert the temperature to kelvins

T = (25.0 + 273.15) K = 298.15 K

(d) Calculate the moles of hydrogen

$\begin{array}{rcl}\text{0.9687 atm}* \text{0.246 L} & = & n * 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{298.15 K}\n0.2383 & = & 24.47n \text{ mol}^(-1)\n\nn & = & \frac{0.2383}{24.47\text{ mol}^(-1)}\n\n& = & 0.009740 \text{ mol}\n\end{array}$

3. Moles of Na

The molar ratio is 2 mol Na: 1 mol H₂

$\text{Moles of Na} =\text{0.009 740 mol H}_(2) * \frac{\text{2 mol Na}}{\text{1 mol H}_(2)} = \text{0.019 48 mol Na}$

4. Mass of Na

$\text{Mass of Na} = \text{0.019 48 mol Na} * \frac{\text{22.99 g Na}}{\text{1 mol Na}} = \text{0.449 g Na}\n\n\text{The mass of Na used was $\large \boxed{\textbf{0.449 g}}$}$

The number of grams of sodium used in the reaction will be approximately 0.1387 grams.

To calculate the number of grams of sodium used in the reaction, we need to use the ideal gas law and consider the effect of the vapor pressure of water on the pressure of the collected hydrogen gas.

Given data:

Volume of hydrogen gas (V) = 246 mL = 0.246 L

Pressure of hydrogen gas (P) = 1.00 atm

Vapor pressure of water (P_water vapor) = 0.0313 atm (subtracted from the total pressure)

Temperature (T) = 25.0°C = 298.15 K

The ideal gas law is given by the equation: PV = nRT, where n is the number of moles of the gas.

First, calculate the total pressure by subtracting the vapor pressure of water from the given pressure of the gas:

Total pressure (P_total) = P - P_water vapor

= 1.00 atm - 0.0313 atm = 0.9687 atm

Now, rearrange the ideal gas law equation to solve for n (moles of gas):

n = PV / RT

Plug in the values:

n = (0.9687 atm × 0.246 L) / (0.0821 L·atm/mol·K × 298.15 K)

n ≈ 0.01206 mol

According to the balanced chemical equation, 2 moles of sodium (Na) produce 1 mole of hydrogen gas (H₂). Therefore, the number of moles of sodium used in the reaction is half of the calculated moles of hydrogen gas:

Moles of sodium = 0.01206 mol / 2 = 0.00603 mol

Finally, calculate the mass of sodium (molar mass of sodium = 22.99 g/mol)

Mass of sodium = Moles of sodium × Molar mass of sodium

Mass of sodium = 0.00603 mol × 22.99 g/mol

≈ 0.1387 g

Therefore, the number of grams of sodium used in the reaction is approximately 0.1387 grams.

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In a coffee-cup calorimeter, 100.0 ml of 1.0 m naoh and 100.0 ml of 1.0 m hcl are mixed. both solutions were originally at 24.68c. after the reaction, the final temperature is 31.38c. assuming that all the solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.18 j/8c ? g, calculate the enthalpy change for the neutralization of hcl by naoh