Answer:
The pH changes by 2.0 if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.
Explanation:
To solve this problem we use the<em> Henderson-Hasselbach equation</em>:
Let's say we have a weak acid whose pKa is 7.0:
If the [A⁻]/[HA] ratio is 10/1, we're left with:
Now if the ratio is 1/10:
The difference in pH from one case to the other is (8.0-6.0) 2.0.
<em>So the pH changes by 2.0</em> if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.
<u>Keep in mind that no matter the value of pKa, the answer to this question will be the same.</u>
Frostbite to the skin and Severe burning to the eyes.
Answer:
Most similar ----- Lithium
Least similar ---- Nitrogen
Explanation:
Cesium is an element on the periodic with the atomic number 133. It lies in group 1 (i.e., the alkali metals) and period 6 on the periodic table. The oxidation state of group 1 metals is +1. Cesium forms an oxide with oxygen as
.
The most similar compound to this chemical compound is Lithium because Lithium happens to be in the same group one metal with Cesium and forms the compound
with the oxygen
The least similar compound nitrogen due to fact that it is an oxide that is covalent in nature and lies between-group 3 -17 to form an
with oxygen.
This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.