<h2>Answer:</h2>
Arrangement of inter molecular forces from strongest to weakest.
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces.
<h3>Explanation:</h3>
Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.
Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.
Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.
But they are weaker than hydrogen bonding.
London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.
Answer: i would say D or the last one.
Explanation: According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.
Hope this helps :) Can u plz mark me branliest
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)
= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant can be calculated as:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
