D. 18 x 10^23 is the total number of atoms in 1.0 mole of CO2
Answer:
An Atom.
Explanation:
An atom is the smallest particle of a chemical element that can exist. It the fundamental piece of matter. Therefore, splitting carbon into smaller pieces , the smallest piece that would still be called "carbon" would be an Atom of carbon. However, atom is also made of three subatomic particles called electrons, protons and neutrons.
Answer : Al
Explanation :
- For a given redox reaction there must be an oxidizing agent/(s) and a reducing agent/(s).
- the oxidizing agent is the substance that undergoes reduction process (gaining electrons or loss of oxygen atoms), meanwhile, the reducing agent is the substance that undergoes oxidation process (loss of electrons or gaining of oxygen atoms).
- In the reaction above, the oxidation number of (Al) in AlCl3 is (3+). However, the oxidation number of (Al) in the products is zero because it exists as a single element.
Therefore, changing from (3+) to zero means gaining of (3) electrons to neutralize the previously existing (3) protons on (Al) in AlCl3.
So Al is the oxidizing agent..
Explanation:
Ionic bonds are bonds formed as a result of the electrostatic attraction between two species.
- This bond type is an interatomic bond.
- It forms when two specie with a large electronegative difference between them combines.
- This is usually a metal and a non-metal.
- The metal loses its valence electrons and becomes positively charged.
- The non-metal gains the electron and becomes negatively charged.
- An electrostatic attraction between the two specie leads to the formation of an ionic bond.
- They are solids with a high melting point.
- They are soluble in polar solvents.
learn more:
Ionic bonds brainly.com/question/6071838
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Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
