Answer:
401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.
Explanation:
The expression for the standard change in free energy is:
Where,
is the change in the Gibbs free energy.
T is the absolute temperature. (T in kelvins)
is the enthalpy change of the reaction.
is the change in entropy.
Given at:-
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
= 128.9 kJ/mol
= 33.1 kJ/mol
Applying in the above equation, we get as:-
= 0.32131 kJ/Kmol
So, For reaction to be spontaneous, 
Thus, For minimum temperature:-
<u>Hence, 401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.</u>
Light years? if you are talking about the distance to the star.
A grey coloured rock with amphibole and intermediate plagioclase like an andesine would classify as an intermediate rock by Bowen's Reaction Series and by the classification of igneous rocks would probably be like a diorite which is intermediate between a gabbro and a granite. A diorite essentially has no quartz but has the silicates amphibole (like hornblende), mica perhaps a little pyroxene and andesine plagioclase.
Answer:
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Explanation:
Step 1: Data given
The initial volume of the sample = 2.00 L
The initial temperature = 35 °C = 308 K
The increased volume = 5.25 L
Pressure = constant
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒ with V1 = the initial volume = 2.00 L
⇒ with T1 = the initial volume = 308 K
⇒ with V2 = the new volume = 5.25 L
⇒ with T2 = the new temperature
2.00 / 308 = 5.25 / T2
0.00649 = 5.25/T2
T2 = 5.25/ 0.00649
T2 = 808.9 K
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
