The only reaction that will most likely take place based on the activity series is; C: Ba + Na₂CO₃
<h3>Activity Series</h3>
From the activity series given, we can say that during a displacement reaction, the more reactive metal will always displace the lesser reactive metal from a compound.
Let us look at each of the given options;
A) Ag + FeCl₃
From the activity series, Silver(Ag) is less reactive than Iron(Fe) and as such Silver cannot displace iron from its' compound.
B) Zn + CaO
Zinc (Zn) is a less reactive metal than Calcium (Ca), so it cannot cannot displace Calcium from its' compound.
C) Ba + Na₂CO₃
Barium (Ba) is a more reactive metal than Sodium (Na), and therefore Barium will displace sodium from its compound in a displacement reaction.
D) Cr + KNO₃
Chromium (Cr) is a lesser reactive metal than Potassium (K), and it therefore cannot displace Potassium from its compound.
Complete question is;
Which reaction will most likely take place based on the activity series? Li > K Ba > CaNa > Mn > Zn > Cr> Fe> Cd Ni > H > Sb > Cu > Ag > Pd > Hg > Pt
A. Ag + FeCl₃
B. Zn + CaO
C. Ba + Na₂CO₃
D. Cr + KNO₃
Read more about Activity Series at; brainly.com/question/17469010
Answer:
In the first equation x=4 or x=-3
In the second equation x=-4 or x=3
Explanation:
The first equation x^2 − x − 12 can be factorized as follows:
x^2-4x+3x-12
x(x-4)+3(x-4)
x-4=0 or x+3=0
x=4 or x=-3
It is noteworthy that -x was rewritten as -4x+3x in order to solve the equation
The second equation x^2 + x − 12 can be factorized as below:
x^2+4x-3x-12
x(x+4)-3(x+4)
x+4=0 or x-3=0
x=-4 or x=3
It is noteworthy that +x was rewritten as 4x-3x in order to solve the equation
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The net electric field is the vector sum of the components of the electric
field produced by the two charges.
The values of the magnitude and direction of the net electric field at the origin (approximate values) are;
- 12.6 ° above the negative x–axis
<h3>How are the net electric field magnitude and direction calculated?</h3>
The possible questions based on a similar question posted online are;
(a) The net electric field at the origin.
The electric field due to charge q₁ is given as follows;
Which gives;
Which gives;
Therefore;
![\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)](https://tex.z-dn.net/?f=%5Cvec%20%20%7BE%7D%20%3D%20%5Cleft%5B%2021.6%20%5C%2C%20N%2FC%20-%20150%20%5C%2C%20N%2FC%20%5Cright%5D%20%5Cleft%28%20%5Chat%20x%20%5Cright%29%20%2B%20%5Cleft%2828.8%20%5C%2C%20N%2FC%20%5Cright%29%20%5Cleft%28%20%5Chat%20y%20%5Cright%29)
The magnitude of the net electric field is therefore;
E = 
≈ 131.6
- The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>
(b) The direction of the net electric field at the origin.
Learn more about electric field strength here:
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