Some students were building models of solids, liquids, and gases by gluing plastic balls to a sheet of cardboard. which of their
models best shows the particles in a solid?
1 answer:
Below picture contains the given models.
Answer:
Option-B is the correct model.
Explanation:
Solid is a state of matter in which the particles are closely packed, has definite volume and shape. Like liquids they don't flow, either they occupy the volume of container as that occupied by both gases and liquids. The inter-molecular forces between solid particles are very strong as compared to liquids ans gases. So, the model B has a particles very closely packed to each other.
Answer:
Option-B is the correct model.
Explanation:
Solid is a state of matter in which the particles are closely packed, has definite volume and shape. Like liquids they don't flow, either they occupy the volume of container as that occupied by both gases and liquids. The inter-molecular forces between solid particles are very strong as compared to liquids ans gases. So, the model B has a particles very closely packed to each other.
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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:
= 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