Answer:
a) MZ₂
b) They have the same concentration
c) 4x10⁻⁴ mol/L
Explanation:
a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.
The salts solubilization reactions and their Kps values are:
MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²
MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³
Thus, the Kps of MZ₂ has a larger value.
b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.
c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:
8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.
Answer:
See explanation
Explanation:
The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.
The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.
Sample means for solutions 1 and 2 are 19.27 and 10.32 respectively
In semiconductor manufacturing,
The total for answer 1 is given by:
9.7+10.5+9.4+10.6+9.3+10.7+9.6+10.4+10.2+10.5 = 192.7
The sample size is 10 and provides us with
192.7/10 = 19.27
For solution 2, the sum is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10, this gives us
103.2/10 = 10.32
The total for answer 2 is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10 and provides us with
103.2/10 = 10.32
Learn more about semiconductor manufacturing here brainly.com/question/22779437
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In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and is known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. Assume the variances are equal. The etch rates are as follows (in mils per minute): Solution 1 Solution 2 9.7 10.6 10.5 10.3 9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4 10.2 10.1 10.5 10.3 Calculate sample means of solution 1 and solution 2
Answer:
cis 2,3-dibromo- 2-butene
trans 2,3-dibromo- 2-butene
Explanation:
The cis-trans or geometric isomerism is due to restricted rotation around a carbon-carbon bond. This restriction may be due to the presence of double bonds or cycles.
The carbon-carbon double bond prevents free rotation of atoms in molecules. These two molecules have the same atoms, but they are different molecules. They are geometric isomers to each other.
The given compound can exist in the form of two isomers, cis and trans. The isomer that has the substituents on the same side is called cis, and the one that has them on opposite sides is trans.
Answer:
E) Two of the above statements are true.
Explanation:
The options are:
A) Before the solution is titrated with HCl it is pink and when the color changes from pink to colorless, the moles of H*(aq) equals the moles of OH"(aq) used in the hydrolysis of the neutralized aspirin. <em>TRUE. </em>Before the solution is titrated, there is an excess of NaOH (Basic solution, phenolphtalein is pink). Then, at equivalence point, after the addition of HCl, the pH is acidic and phenolphtalein is colorless.
B) Before the solution is titrated with HCl it is colorless and when the color changes from colorless to pink, the moles of H*(aq) equals the excess moles of OH(aq) added. <em>FALSE. </em>As was explained, before the titration, the solution is pink.
C) 25.0 mL of 0.100 M NaOH was added to the sample to hydrolyze the neutralized aspirin in the solution. The titration with HCl allows us to determine the moles of excess OH(aq) added. Once we determine the moles of excess OH(aq), we can determine moles of OH"(aq) used in the hydrolysis of the neutralized aspirin, which is equal to the moles of aspirin in the recrystallized aspirin. <em>TRUE. </em>Aspirin requires an excess of base (NaOH) for a complete dissolution (Hydrolysis). Then, we add H+ as HCl to know the excess moles of OH-. As we know the added moles of OH-, we can find the moles of OH that reacted = Moles of aspirin.
D) We can determine the moles of aspirin in the recrystallized aspirin by titrating with the 0.100 M NaOH to the neutralization point. The purpose of the hydrolysis of the neutralized aspirin and the back-titration with the 0.100 M HCl is to confirm the moles of aspirin in the recrystallized aspirin. <em>FALSE. </em>NaOH can be added directly unyil neutralization point because, initially, aspirin can't be dissolved completely
E) Two of the above statements are true. <em>TRUE</em>
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Right option is:
<h3>E) Two of the above statements are true.</h3>
