M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL
From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
Answer:
The rate law for the decomposition reaction is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be 
Explanation:

The rate law can be expressed as:
..[1]
When the drug concentration is tripled, the decomposition rate rises by a factor of nine.
![[D]'=3[D]](https://tex.z-dn.net/?f=%5BD%5D%27%3D3%5BD%5D)

...[2]
[1] ÷ [2]
![\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR%27%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5BD%27%5D%5Ex%7D)
![\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B9R%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5B3D%5D%5Ex%7D)

Solving for x results in:
x = 2.
This indicates a second-order reaction.
The decomposition reaction's rate law is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be:
![k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BR%7D%7B%5BD%5D%5E2%7D%3D%5Cfrac%7BM%2Fs%7D%7B%28M%29%5E2%7D%3DM%5E%7B-1%7Ds%5E%7B-1%7D)
The unit for the rate constant will be
.
Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO
Answer: The right choice is (c) application of both a mobile phase and a stationary phase.
Explanation:
Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.
Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.
Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.
Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.
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.