Answer:
The median would be 6700
Explanation:
Arrange data values from lowest to highest value
The median is the data value in the middle of the set
.
Ordering a data set x1 ≤ x2 ≤ x3 ≤ ... ≤ xn from lowest to highest value, the median x˜ is the data point separating the upper half of the data values from the lower half.
If the size of the data set n is odd the median is the value at position p where
Formula for the median
p=n+12
x˜=xp
If n is even the median is the average of the values at positions p and p + 1 where
p=n2
x˜=xp+xp+12
If there are 2 data values in the middle the median is the mean of those 2 values.
Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se
the reason is P dopant will introduce holes in the Se as P has lesser valence electron
Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
Concentration is decreased to 1.56 % which means that 0.0156 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 0.0156
Thus,
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,
<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>
Answer:
The correct answer is b. 1280 cm^2
