Li+ has a smaller ionic radius than K+
and smaller molecules have more collisions/interactions between each other
<h3>What is ion-solvent interaction ?</h3>
In the case of ion-solvent interactions, the state in which the interac-tions exist is an obvious one; it is the situation in which ions are inside the solvent.
- Ions are charged particles, and charges interact with other charges. So there will also be ion-ion, as well as ion-solvent, interactions in the solution.
- In the process of solvation, ions are surrounded by a concentric shell of solvent. Solvation is the process of reorganizing solvent and solute molecules into solvation complexes.
Learn more about Ion-solvent interaction here:
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Answer:
116 g
Explanation:
From the question given above, the following data were obtained:
Number of mole of calcium = 2.9 moles
Mass of calcium =.?
The mole and mass of a substance are related according to the following formula:
Mole = mass / molar mass
With the above formula, we can obtain the mass of calcium. This can be obtained as follow:
Number of mole of calcium = 2.9 moles
Molar mass of calcium = 40 g/mol
Mass of calcium =.?
Mole = mass / molar mass
2.9 = mass of calcium / 40
Cross multiply
Mass of calcium = 2.9 × 40
Mass of calcium = 116 g
Therefore, the mass of 2.9 moles of calcium is 116 g.
Answer:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
Answer:
The answer to your question is: 6.55 x 10 ²³ atoms of Br
Explanation:
CH2Br2 = 37.9 g
MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g
174 g of CH2Br2 ------------------ 160 g of Br2
37.9 g of CH2Br2 --------------- x
x = 37.9 x 160/174 = 34.85 g of Br
1 mol of Br ----------------- 160 g Br2
x ---------------- 174 g Be2
x = 174 x 1 /160 = 1.088 mol of Br2
1 mol of Br ----------------- 6.023 x 10 ²³ atoms
1.088 mol of Br ------------- x
x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms
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