Ionic compounds generally occur between metals and non-metals due to their large electronegativity difference. You can simple go down Group 1 and Group 17 of the periodic table.
Examples:
NaCl (Sodium Chloride)
KCl (Potassium Chloride)
RbCl (Rubidium Chloride)
CsCl (Cesium Chloride)
The molarity of the stock Mn²⁺ ions is 0.0288 M
Based on the dilution formula;
- The molarity of A is 0.00144 M
- The molarity of B is 0.0000576 M
- The molarity of C is 0.000001152 M
<h3>What is the molarity of a solution?</h3>
The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.
- Molarity = number of moles/volume
The molarity of the stock solution is:
moles of Mn²⁺ ions = mass / molar mass
molar mass of Mn²⁺ ions = 55.0 g/mol
moles of Mn²⁺ ions = 1.584 / 55
moles of Mn²⁺ ions = 0.0288 moles
molarity of Mn²⁺ ions = 0.0288 / 1
molarity of Mn²⁺ ions = 0.0288 M
The dilution formula is used to determine the molarities of A, B, and C.
C₁V₁ = C₂V₂
C₂ = C₁V₁ / V₂
Where;
- C₁ = initial molarity
- V₁ = initial volume
- C₂ = final molarity
- V₂ = final volume
Molarity of A = 50 * 0.0288 / 1000
Molarity of A = 0.00144 M
Molarity of B = 10 * 0.00144 / 250
Molarity of B = 0.0000576 M
Molarity of C = 10 * 0.0000576 / 500
Molarity of C = 0.000001152 M
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The number of mole of nitrogen that occupies 1.2 L under the same condition is 0.6 mole
<h3>Data obtained from the question </h3>
- Initial mole (n₁) = 0.2 mole
- Initial volume (V₁) = 0.4 L
- Final volume (V₂) = 1.2 L
- Final mole (n₂) =?
<h3>How to determine the final mole </h3>
The final mole can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side n
PV / n = RT
Divide both side by P
V / n = RT / P
RT / P = constant
V / n = constant
Thus,
V₁ / n₁ = V₂ / n₂
0.4 / 0.2 = 1.2 / n₂
2 = 1.2 / n₂
Cross multiply
2 × n₂ = 1.2
Divide both side by 2
n₂ = 1.2 / 2
n₂ = 0.6 mole
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T₁ = 50,14 K.
p₁ = 258,9 torr.
T₂ = 161,2 K.
p₂ = 277,5 torr.
R = 8,314 J/K·mol.
Using Clausius-Clapeyron equation:
ln(p₁/p₂) = - ΔHvap/R · (1/T₁ - 1/T₂).
ln(258,9 torr/277,5 torr) = -ΔHvap/8,314 J/K·mol · (1/50,14 K - 1/161,2 K).
-0,069 = -ΔHvap/8,314 J/K·mol · (0,0199 1/K - 0,0062 1/K).
0,0137·ΔHvap = 0,573 J/mol.
ΔHvap = 41,82 J.
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