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Harlamova29_29 [7]
3 years ago
8

Water moves to the top of tall trees because of

Chemistry
1 answer:
Fynjy0 [20]3 years ago
4 0
Capillary Action ( combination of cohesion and adhesion )
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Colligative properties depend most on which of the following?
erma4kov [3.2K]

Answer: -

Thus Colligative properties depend most on concentration of solute particles.

Explanation: -

Colligative properties are the properties of a solution that depends only on the number of particles of the solute present in the solution and not on the identity of the solute.

The number of solute particles present per unit volume in a solution is referred to as the concentration of solute particles.

Thus Colligative properties depend most on concentration of solute particles.

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3 years ago
Can someone help me with my chemistry on my page/account?
lesantik [10]

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Yes

Explanation:

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3 years ago
What happens if a species is unable to adapt to a changing environment.
enot [183]

The awnser your looking for would be B: It's Population decreases. Because if a species is unable to adapt to a changing environment the species would either have to move away or they could die in that location.


4 0
3 years ago
If two protons and two neutrons are added to the nucleus of a carbon atom, what nucleus does it become?
Shkiper50 [21]
If two protons are added to carbon, the chemical identity changes to oxygen, which is two spaces over horizontally on the periodic table. The number of protons is what we call the atomic number (Z), and this is what defines the identity of an element. Since we also added two neutrons, this is simply oxygen, or oxygen-16, the most abundant isotope of oxygen.
5 0
3 years ago
Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper
KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

  • CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0
3 years ago
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