Answer:
k = 0.0306 min-1
Explanation:
The table is given as;
Time, Concentration
0 1.48
5 1.27
10 0.98
15 0.84
The integrated rate law for a first order reaction is given as;
ln [A] = -kt + ln [Ao]
where;
[A] = Final Concentration
[Ao] = Initial Concentration
k = rate constant
t = time
In the table, taking the first two sets of values;
t = 5
k = ?
[Ao] = 1.48
[A] = 1.27
Inserting into the equation;
ln(1.27) = - k (5) + ln(1.48)
ln(1.27) - ln(1.48) = -5k
-0.1530 = -5k
k = -0.1530 / -5
k = 0.0306 min-1
Those elements with similar properties are in the same column.
The molar Concentration of KMnO₄ is 0.000219 M
Concentration is the abundance of a constituent divided by means of the overall extent of an aggregate. numerous styles of mathematical description may be outstanding: mass awareness, molar awareness, variety concentration, and quantity awareness.
y is absorbance
x is the molar concentration of KMnO_4
y = 4.84E + 03x - 2.26E - 01
0.833 = 4.84 * 10⁺⁰³ x - 2.26 * 10⁻¹
1.059 = 4.84 * 10⁺⁰³ x
X = 0.000219 M
Hence, The molar Concentration of KMnO₄ is 0.000219 M
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Half life is the time taken by a radioactive isotope to decay by half its original mass. In this case, the halflife of the radioactive isotope is 5000 years.
Initially the mass is 100 %; thus the mass that will be left will be given by;
New mass = Original mass × (1/2)^n where n is the number of half lives;
n = 10000/5000 = 2
New mass = 100% ×(1/2)^2
= 100 % × 1/4
= 25%
Therefore; the mass left after 10000 years is 25% or 1/4 of the original mass.
For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
