Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
1 answer:
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Answer:
Explanation:
Diethyl malonate possesses greater acidity compared to monocarbonyl substances (pKa=13) because its alpha hydrogens are linked to two carbonyl groups. Consequently, the malonic ester can be readily changed into its enolate ion by reacting it with sodium ethoxide in ethanol. When the malonic ester undergoes alkylation, a hydrogen atom in the alpha position becomes acidic, permitting another round of alkylation to yield a dialkylated malonic ester.
In this scenario, when diethyl malonate interacts with urea in the presence of sodium ethoxide base, the second alkylation step occurs within the molecule, producing a cyclic compound known as barbituric acid.
Answer:
1) This dilution plan will yield a 200μM solution.
2) This dilution plan will not yield a 200μM solution.
3) This dilution plan will not yield a 200μM solution.
4) This dilution plan will yield a 200μM solution.
5) This dilution plan will yield a 200μM solution.
Explanation:
Convert the initial molarity into molar form as shown.
Let's examine the following serial dilution processes.
1)
Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.
Concentration of 500 mL solution:
10 mL of this solution is further diluted to 250 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.
2)
Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.
Concentration of 100 mL solution:
10 mL of this solution is further diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will not yield a 200μM solution.
3)
Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.
Concentration of 100 mL solution:
5 mL of this solution is diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will not yield a 200μM solution.
4)
Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.
Concentration of 250 mL solution:
10 mL of this solution is further diluted to 500 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.
5)
Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.
Concentration of 250 mL solution:
10 mL of this solution is further diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.
Greetings!
The result is:
The new volume is:
Rationale:
Because the temperature remains constant, we can apply Boyle's Law to solve this issue.
Boyle's Law stipulates that:
Where,
P is the gas's pressure.
V is the gas's volume.
According to the information provided:
Let's put the values into the equation:
Consequently, the new volume is:
Wishing you a lovely day!
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.
