A bowling ball.
definition of dense-closely compacted in substance
Answer:
473 year
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
To reach 12.5% of reactant means that 0.125 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 0.125
t = ?
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
t = 473 year
Answer:
ΔG = -52.9 kJ/mol
Explanation:
Step 1: Data given
Temperature = 298 K
All species have a partial pressure of 1 atm
Δ G ° = − 69.0 kJ/mol
Step 2: The balanced equation
N2(g) + 3H2(g) ⇆ 2NH3 (g)
Step 3: Calculate Q
we will use the expression: ΔG = ΔG° + RT*ln(Q)
⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67
Step 4: Calculate ΔG
So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol
(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)
Answer:
Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature.
Explanation:
