Interstellar dust particles help cool the nebular cloud by _____.
1 answer:
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Answer:
Becomes 1/4 as much
Explanation:
If a and b are directly proportional, this means we may write an equality between them if some proportionality constant, k, is included, so that:
The equation shows that:
And since k is a constant, this would imply that increasing a by some fraction would also require b to increase by the same fraction.
In the context of this problem, we're given that a becomes 1/4 as much, meaning b should also become 1/4 as much to keep the same proportionality constant, k.
The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ
<h3>Further explanation</h3>Enthalpy is the amount of system heat at constant pressure.
The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The standard unit is kilojoules (kJ)
The enthalpy change symbol (ΔH) is usually written behind the reaction equation.
Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.
Examples of water evaporation:
H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ
The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid
- 155 g of butane
relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol
tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]
mole = 2,672
Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:
23.1 kJ / mol x 2,672 mol = 61,723 kJ
<h3>Learn more</h3>the heat of vaporization
The latent heat of vaporization
Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic
Answer:
Glycogen. Cellulose. Amylose. Cellulose. Amylopetin and Glycogen. Amylopetin and Cellulose.
Explanation:
Glycogen is the form that glucose is stored in human body.
Cellulose is the structural part of plant cell walls and human cannot digest it.
Amylose is the polysaccharide linked mainly by the the bonds of 1,4 glycosidic.
Cellulose is an unbranched polysaccharide linked mainly by the bonds of 1,4 glycosidic.
Amylopetin and Glycogen are branched polysaccharides linked by the bonds of 1,4 glycosidic and
1,6 glycosidic.
Amylopetin and Cellulose are mainly stored in plants.