Answer:
The cathode of a spontaneous electrochemical cell is the cell in which metal cations undergo reduction. The electrons are supplied by the anode. Without much thought, we can conclude magnesium is much less favorably reduced than gold.
Answer:
A. Interactions between the ions of sodium chloride (solute-solute interactions).
B. Interactions involving dipole-dipole attractions (solvent-solvent interactions).
C. Interactions formed during hydration (solute-solvent interactions).
D. Interactions involving ion-ion attractions (solute-solute interactions).
E. Interactions associated with an exothermic process during the dissolution of sodium chloride (solute-solvent interactions).
F. Interactions between the water molecules (solvent-solvent interactions).
G. Interactions formed between the sodium ions and the oxygen atoms of water molecules (solute-solvent interactions).
Explanation:
The solution process takes place in three distinct steps:
- Step 1 is the <u>separation of solvent molecules.
</u>
- Step 2 entails the <u>separation of solute molecules.</u>
These steps require energy input to break attractive intermolecular forces; therefore, <u>they are endothermic</u>.
- Step 3 refers to the <u>mixing of solvent and solute molecules.</u> This process can be <u>exothermic or endothermic</u>.
If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (ΔHsoln < 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (ΔHsoln > 0).
In the dissolution of sodium chloride, this process is exothermic.
CaCO₃ → CaCl₂
Ca = 40, C = 12, O = 16, Cl = 35.5
100 →111
3 → x
x = 3.33g
Answer:
Explanation:
We will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 315.46 76.12
Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O
m/g: 6.5
1. Moles of Ba(OH)₂·8H₂O
2. Moles of NH₄SCN
The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O
3. Mass of NH₄SCN
Answer:
3.1% is the fraction of the sample after 28650 years
Explanation:
The isotope decay follows the equation:
Ln[A] = -kt + Ln[A]₀
<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>
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k could be obtained from Half-Life as follows:
K = Ln 2 / Half-life
K = ln 2 / 5730 years
K = 1.2097x10⁻⁴ years⁻¹
Replacing in isotope decay equation:
Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]
Ln[A] = -3.4657
[A] = 0.0313 =
<h3>3.1% is the fraction of the sample after 28650 years</h3>
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