Warm is less heaver then cold air so warm air rise and cold air sinks
Answer:
x = 1, -7.5
Explanation:
2x² + 13x = 15
Divide the left side of the equation by 2
2(x² + 6.5x) = 15
Divide 6.5 by 2 and square the quotient
6.5/2 = 3.25
3.25² = 10.5625
Add 10.5625 to the left side
2(x² + 6.5x + 10.5625) = 15
Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.
10.5625 • 2 = 21.125
2(x² + 6.5x + 10.5625) = 36.125
To factor (x² + 6.5x + 10.5625), add b/2 to x
b/2 = 6.5/2 = 3.25
2(x + 3.25)² = 36.125
Divide by 2
(x + 3.25)² = 18.0625
Square root.
(x + 3.25) = √18.0625
x + 3.25 = ±4.25
Subtract 3.25.
x = 4.25 - 3.25 = 1
x = -4.25 - 3.25 = -7.5
x = 1, -7.5
Answer:
An ion is defined as an atom or molecule that has gained or lost one or more of its valence electrons, giving it a net positive or negative electrical charge.
Explanation:
I am here zinda -_+...
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Either it’s, it is released when the reaction is complete or it is changed into atoms of carbon and oxygen during the reaction