Answer:
9 (1-2x²)
Explanation:
The given expression is:
30 - 9x²*2 - 21 - 4 + 4
The first step is to compute the multiplication. This will give:
30 - 18x² - 21 - 4 + 4
Then, we will add like terms as follows:
(30-21-4+4) - 18x²
= 9 - 18x²
Finally, we can take the 9 as a common factor from both terms, this will give:
9 (1-2x²)
Hope this helps :)
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped :D
Answer:
A) t = 22.5 min and B) t = 29.94 min
Explanation:
Initial concentration, [A]₀ = 100
Final concentration = 100 -75 = 25
Time = 45 min
A) First order reaction
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[25] − ln[100] = - 45k
-1.386 = -45k
k = 0.0308 min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[50] − ln[100] = - 0.0308 * t
-0.693 = -0.0308 * t
t = 22.5 min
B) Zero Order
[A] = [A]₀ − kt
Using the values from the initial reaction and solving for k, we have;
25 = 100 - k(45)
-75 = -45k
k = 1.67 M min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
[A] = [A]₀ − kt
50 = 100 - (1.67)t
-50 = - 1.67t
t = 29.94 min
a) The change here is that metallic iron is converted into ions and copper is deposited. This is called a displacement reaction.
b) 
is oxidised in this reaction.
c)$ \mathrm{Fe_{(s)}+ CuSO_{4(aq)} \rightarrow FeSO_{4(aq)} + Cu_{(s)}}$
Answer:
False
Explanation:
The octet rule forms the basis for chemical reactions. The octet rule states that; an atom is only stable when it has eight electrons around its outermost shell.
This implies that the driving force behind chemical reaction is the attainment of an octet structure(eight electrons in the outermost shell of each of the bonding atoms).
An atom that has only six electrons in its outermost shell is not yet stable according to the demand of the octet rule. Hence, the statement "chemical reactions happen and compounds form because they're trying to get 6 electrons in their outer orbitals" is false.
