Answer:
Explanation:
Hello there!
In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:
Thus, we solve for the initial volume, V1, as shown below:
And plug in the initial concentration and final concentration and volume to obtain:
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Answer:
CaCl₂ > CH₃OH = LiCl > C₆H₁₂O₆
Explanation:
The osmotic pressure of a compound is calculated using the following expression:
π = MRT (1)
This expression is used when the substance is nonelectrolyte. If the solution is electrolyte solution then we need to count the van't hoff factor into the expression so:
π = MRTi (2)
Now, we have 4 solutions here, only two of them are electrolyte solution, this means that these solutions can be dissociated into separate ions. These solutions are LiCl and CaCl₂. It can be shown in the following reactions:
LiCl -------> Li⁺ + Cl⁻ 2 ions (i = 2)
CaCl₂ ---------> Ca²⁺ + 2Cl⁻ 3 ions (i = 3)
The methanol (CH₃OH) and glucose (C₆H₁₂O₆) are non electrolyte solutions, therefore they are not dissociated. So, let's use expression (1) for methanol and glucose, and expression (2) for the salts:
CaCl₂: π = 1 * 3 * RT = 3RT
CH₃OH: π = 2 * RT = 2RT
C₆H₁₂O₆: π = 1 * RT = 1RT
LiCl: π = 1 * 2 * RT = 2RT
Finally with these results we can conclude that the decreasing order of these solutions according to their osmotic pressures are:
<h2>
CaCl₂ > CH₃OH = LiCl > C₆H₁₂O₆</h2>
Answer:
1.3 × 10³ g
Explanation:
Step 1: Convert the mass of calaverite to grams
We will use the relationship 1 kg = 1,000 g.
Step 2: Establish the appropriate mass ratio
The mass ratio of AuTe₂ to Au is 452.17:196.97.
Step 3: Calculate the maximum mass of pure gold that can be obtained from 3.0 × 10³ g of calaverite
Answer:
H2TeO3
HClO2
HClO
Explanation:
Ka is the value of the equilibrium constant of the acids. As higher is it's valued, as higher is the value of the concentration of H+, so more acidic is the solution. Another way to determine the acidic is to obtain the pKa = -logKa, in this case, as larger the pKa, as less acidic is the solution. The values of pKa can be found at analytic tables.
First row:
H2, pKa = 15.74
TeO4, pKa = 11
H2TeO3, pKa1 = 2.7
More acidic = H2TeO3
Second row:
HClO2, pKa = 1.94
HClO, pKa = 7.40
More acidic = HClO2
Third row:
HClO, pKa = 7.40
HBrO, pKa = 8.55
More acidic = HClO