Answer:Crop rotation also helps to battle against the forces of erosion. Rotating crops helps to improve soil stability by alternating between crops with deep roots and those with shallow roots. Pests are also deterred by eliminating their food source on a regular basis.
Crop rotation is the practice of growing a series of dissimilar or different types of crops in the same area in sequenced seasons. It is done so that the soil of farms is not used for only one set of nutrients. It helps in reducing soil erosion and increases soil fertility and crop yield.
The goal of crop rotation is to reduce the amount of the pest population present in the soil. Some pathogens that cause diseases survive in the soil from year to year in one form or the other, usually as sclerotia, spores, or hyphae. Rotating to non-host crops prevents the buildup of large populations of pathogens.
Explanation: the answer would be 4
Increases soil fertility. ...
Increases crop yield. ...
Increase in soil nutrients. ...
Reduces soil erosion. ...
Limits concentration of pests and diseases. ...
Reduces the stress of weeds. ...
Improves the soil structure. ...
Reduces pollution.
Answer:
I believe the first one is correct but I think the second one would be the blood like substance. not a 100% sure sorry
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)


Given s = 7.4×10⁻³ M
So, Ksp is:


Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)

Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
Answer:
it is ammonia nitro oxide