Answer:
Ka = 1.78 × 10⁻⁵
Explanation:
Let's consider the following thermochemical equation.
CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq) ∆G° = 27.1 kJ/mol
At 25°C (298 K), we can find the equilibrium constant (Ka) using the following expression.
∆G° = - R × T × lnKa
where,
R: ideal gas constant
T: absolute temperature
27.1 × 10³ J/mol = - (8.314 J/K.mol) × 298 K × lnKa
Ka = 1.78 × 10⁻⁵
Answer:
30.0 L.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution = (MV)after dilution</em>
M before dilution = 5.0 M, V before dilution = 3.0 L.
M after dilution = 0.5 M, V after dilution = ??? L.
∵ (MV)before dilution = (MV)after dilution
∴ (5.0 M)(3.0 L) = (0.5 M)(V after dilution)
<em>∴ V after dilution = (5.0 M)(3.0 L)/(0.5 M) = 30.0 L.</em>
Answer:
The correct answer is "11.44 ml".
Explanation:
Molarity,
= 0.2 M
Density,
= 1.05 g/ml
Volume,
= 1 L
As we know,
⇒
or,
⇒
On putting the values, we get
⇒
⇒
Now,
⇒
or,
⇒
⇒
⇒
hence,
⇒
or,
⇒
⇒
⇒
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