Answer:
161376 kJ
Explanation:
The equation for standard Gibbs energy free energy for a reaction is -
ΔG⁰ = ΔH⁰ - TΔS⁰
The values of ΔH⁰, T and ΔS⁰ are given in the question as
ΔH⁰ = -2256 kJ
T = 298 K
ΔS⁰ = -549.1 kJ
So, putting these values in the above equation we get
ΔG⁰ = -2256 - (-549.1 × 298)
ΔG⁰ = -2256 + 163631.8 = 161375.8 ≈ 161376 kJ
Thus, the standard Gibbs free energy for the reaction mentioned in the question will be 161376 kJ.
Forces are exerted I believe : all of the above
The action force might be Tyler throwing the ball
I don't know the last one
Explanation:
Current is measured in terms of Ampere abbreviated by (I)
Is the quantity of charge passing in given point in a circuit per unit time
Answer:
The acceleration of the electron is 1.457 x 10¹⁵ m/s².
Explanation:
Given;
initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s
distance traveled by the electron, d = 0.01 m
final velocity of the electron, v = 5.4 x 10⁶ m/s
The acceleration of the electron is calculated as;
v² = u² + 2ad
(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a
(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²
(2 x 0.01)a = 2.91375 x 10¹³
![a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B2.91375%20%5C%20%5Ctimes%20%5C%2010%5E%7B13%7D%7D%7B2%20%5C%20%5Ctimes%20%5C%200.01%7D%20%5C%5C%5C%5Ca%20%3D%201.457%20%5C%20%5Ctimes%20%5C%2010%5E%7B15%7D%20%5C%20m%2Fs%5E2)
Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².