Answer:
option D
Explanation:
first quantum number of a 2s2
n = 2
second quantum number of a 2s2
l = 0 to n-1
l = 0, 1
For s orbital
l = 0
third quantum number of a 2s2
ml = - l to +l
ml = 0
fourth quantum number of a 2s2 electron be
?½ = spin down
½ = spin up
2s2 has 2 electrons 1 spin down and 1 spin up (starts with spin down)
ms = +1/2
Answer:
c is right for this question
Explanation:
Answer:
D: 60 grams.
Explanation:
If you look at the x-axis and where it says 50 degrees Celsius, you can see that it takes 20 grams of solute to be saturated at 100 GRAMS of solvent. However, the question is asking for the amount of solute to saturate a 300 GRAM solution.
This means you will have to multiply the amount of solute needed at 100 grams by 3 to get you to the amount needed for 300 grams.
20 times 3 will get you to your answer, 60 grams.
Answer:
the water and alcohol interact, which means the water doesn't even dissolve the sugar or color as well as it normally would. Oil molecules are not polar so they cannot dissolve either the coloring or the sugar.
Explanation:
Answer:
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.
Explanation:
The unit of k is s⁻¹.
The order of the reaction = first order.
First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
[A] = the concentration of the reactant at time t
k= rate constant
t= time
Here k= 4.70×10⁻³ s⁻¹
t= 4.00
[A₀] = initial concentration of reactant = 0.700 M
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
![\Rightarrow -\frac{d[A]}{[A]}=kdt](https://tex.z-dn.net/?f=%5CRightarrow%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3Dkdt)
Integrating both sides
![\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt](https://tex.z-dn.net/?f=%5CRightarrow%5Cint%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3D%5Cint%20kdt)
⇒ -ln[A] = kt +c
When t=0 , [A] =[A₀]
-ln[A₀] = k.0 + c
⇒c= -ln[A₀]
Therefore
-ln[A] = kt - ln[A₀]
Putting the value of k, [A₀] and t
- ln[A] =4.70×10⁻³×4 -ln (0.70)
⇒-ln[A]= 0.375
⇒[A] = 0.686
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.