Answer:
A. Interactions between the ions of sodium chloride (solute-solute interactions).
B. Interactions involving dipole-dipole attractions (solvent-solvent interactions).
C. Interactions formed during hydration (solute-solvent interactions).
D. Interactions involving ion-ion attractions (solute-solute interactions).
E. Interactions associated with an exothermic process during the dissolution of sodium chloride (solute-solvent interactions).
F. Interactions between the water molecules (solvent-solvent interactions).
G. Interactions formed between the sodium ions and the oxygen atoms of water molecules (solute-solvent interactions).
Explanation:
The solution process takes place in three distinct steps:
- Step 1 is the <u>separation of solvent molecules.
</u>
- Step 2 entails the <u>separation of solute molecules.</u>
These steps require energy input to break attractive intermolecular forces; therefore, <u>they are endothermic</u>.
- Step 3 refers to the <u>mixing of solvent and solute molecules.</u> This process can be <u>exothermic or endothermic</u>.
If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (ΔHsoln < 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (ΔHsoln > 0).
In the dissolution of sodium chloride, this process is exothermic.
You will need the Gas Law:
pV = nRT
Since T and p are constant, R is constant too, then moles increases->volume will increase with the same ratio too!
Answer:
17
Explanation:
the mass number of the element which is protons and neutrons is 17+18=35
electronic configuration is 2, 8, 7
valency of the element is 1 therefore the atomic number is 17
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer: through energy carriers, ATP and NADPH
Explanation:in the light dependent stage,energy from a light photon is used to create ATP through ADP and an inorganic phosphate.
It does this by the transfer of energetic electron from one electron carrier to another.NADPH is also formed.
In the light independent reaction,ATP and NADP are used to reduce carbon dioxide to 3-phosphoglycerate
