A particular substance has a measured solubility in water of 3.6 g/1.0 L. How much of the substance can be dissolved in 650 L of
1 answer:
Answer : 2.34 kg or 2340 g can be dissolved in the given amount of water
The solubility of the substance is given as 3.6 g/1.0 L. That means 3.6 grams of the substance can be dissolved in 1 L of water.
We have 650 L of water. The equation can be set up as follows.
We can cancel out L here
Therefore we can say that 2340 grams or 2.34 kg of the substance can be dissolved in the given amount of water
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Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0
Answer ; The question is missing in some details, but here are he details ;
The two naturally occurring isotopes of bromine are
81Br (80.916 amu, 49.31%) and
79Br (78.918 amu, 50.69%).
The two naturally occurring isotopes of chlorine are
37Cl (36.966 amu, 24.23%) and
35Cl (34.969 amu, 75.77%).
Bromine and chlorine combine to form bromine monochloride, BrCl.
Explanation:
The detaile calculation is as shown in the attachment.
