Answer:
Is B, it is B because you don't need arrows on it.
Answer: A plot of the natural log of the concentration of the reactant as a function of time is linear.
Explanation:
Since it was explicitly stated in the question that the half life is independent of the initial concentration of the reactant then the third option must necessarily be false. Also, the plot of the natural logarithm of the concentration of reactant against time for a first order reaction is linear. In a first order reaction, the half life is independent of the initial concentration of the reactant. Hence the answer.
Answer:
<h2>0.52 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
volume = final volume of water - initial volume of water
volume = 35 - 8 = 27 mL
We have

We have the final answer as
<h3>0.52 g/mL</h3>
Hope this helps you
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = 
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer: true
Explanation: Electrons orbit the nucleus of an atom the way that planets revolve around the Sun. The electrons are like the planets in the solar system. The sun is like the nucleus in the solar system. The answer to the question is true.