The answer is Conduction. I hope this helps!
Element 6 has the lowest ionization energy
Answer:
No, the experimental result is different from the theoretical value.
Explanation:
Based on the given information, the mass of beaker and watchglass plus alum hydrate is 102.218 grams, and the mass of beaker and watchglass is 101.286 grams. Therefore, the mass of alum hydrate is:
= 102.218 grams - 101.286 grams
= 0.932 grams
Now the mass of anhydrous compound is,
= 102.218 grams - 101.798 grams
= 0.42 grams
Thus, the mass of water present is,
= 0.932 grams - 0.42 grams
= 0.512 grams
The mass percent of water is,
= mass of water/Total mass of hydrate * 100
= 0.512 grams / 0.932 grams * 100
= 54.93 %
Hence, the experimental result in not similar to the theoretical result.
The Cambrian period, part of the Paleozoic era, produced the most intense burst of evolution ever known. The Cambrian Explosion saw an incredible diversity of life emerge, including many major animal groups alive today. Among them were the chordates, to which vertebrates (animals with backbones) such as humans belong. These included brachiopods, which lived in shells resembling those of clams or cockles, and animals with jointed, external skeletons known as arthropods—the ancestors of insects, spiders, and crustaceans. These toughened-up creatures represented a crucial innovation: hard bodies offering animals both a defense against enemies and a framework for supporting bigger body sizes.
Answer:
ΔH°r = -184.6 kJ
Explanation:
Let's consider the following balanced equation.
H₂(g) + Cl₂(g) ⇄ 2 HCl(g)
We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression:
ΔH°r = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(p) are the standard enthalpies of formation of reactants and products
By definition, the standard enthalpy of formation of a simple substance in its most stable state is zero. Then,
ΔH°r = 2 mol × ΔH°f(HCl(g)) - [1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(Cl₂(g))]
ΔH°r = 2 mol × (-92.3 kJ/mol) - [1 mol × 0 + 1 mol × 0]
ΔH°r = -184.6 kJ
