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What is the main type of weathering in deserts? weathering by organic acids chemical weathering physical weathering

Solnce55 [7]

Answer:

chemical

Explanation:

Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.

5 0

3 years ago

Calculate the net force 175 N 1125 N

Ket [755]

Answer:

<h3>_____________&_________</h3><h2>Net force = sum of all forces with directions</h2><h2>so.....</h2><h2>NET FORCE = 125+(-75)</h2><h2> = 50 N</h2><h2> in the direction of yellow </h2><h2> t-shirt man</h2><h2>------------------------------------</h2>

4 0

3 years ago

32P can be used to make any nucleotide (A, C, G, or T) radioactive. Which of the following explains why this is true?

xxMikexx [17]

B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.

Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P).

3 0

2 years ago

Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor

sukhopar [10]

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

42.2 mmHg - P' = 42.2 mmHg . 0.0172

P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

P' = 41.4 mmHg

5 0

3 years ago

He isotope 62(over28ni has the largest binding energy per nucleon of any isotope. calculate this value from the atomic mass of n

irina [24]

The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 = 0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.

4 0

3 years ago