Ask question

Ask question

All categories

3 years ago

14

Which of the following is a farming method that helps reduce the effects of wind erosion?

2 answers:

stiks02 [169]3 years ago

6 0

The answer is C.) Placing Wind Barriers

The reasoning is because irrigation, if used too much, causes water erosion. Planting crops deeper in the ground would would prevent the plants from being uprooted. Finally, terracing would prevent from run-off eroding the ground away from the plants.

Tamiku [17]3 years ago

6 0

(c) planting wind barries

You might be interested in

When might neutralization reactions be used in a laboratory setting?

Ipatiy [6.2K]

Neutralization reactions can be used in a laboratory setting in order t<span>o dispose of chemicals. When spills happens, for instance an acid is on the floor, you can use a base to neutralize the spill. Hope this answers the question. Have a nice day.</span>

6 0

2 years ago

The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However

Murljashka [212]

Answer:

Reverse the $2 NO_2 \longrightarrow 2 NO + O_2$ reaction

Explanation:

Reactions:

$2 NO_2 \longrightarrow 2 NO + O_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Overall:

$N_2O_4 \longrightarrow 2 NO_2$

As can be seen, in the overall reaction we have $N_2O_4$ in the reactants like in the second reaction and $NO_2$ in the products. The $NO_2$ is in the first reaction but as a reactant so we need to reverse that reaction:

$2 NO + O_2 \longrightarrow 2 NO_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Combining:

$N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2$

$N_2O_4 \longrightarrow 2 NO_2$

4 0

2 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

2 years ago

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

ch4aika [34]

<u>Answer:</u> The molecular weight of protein is $1.14\times 10^2g/mol$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of protein = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)

$M_{solute}$ = molar mass of protein = ?

$V_{solution}$ = Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol$

Hence, the molecular weight of protein is $1.14\times 10^2g/mol$

4 0

2 years ago

The jet stream is one of the key winds that moves air masses in which direction does the jet stream in the United

Yuri [45]

Answer: West to East

3 0

3 years ago

Read 2 more answers