Answer:
=16.49 L
Explanation:
Using the equation
P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273
P1V1/T1= P2V2/T2
0.6×30/298= 1×V2/273
V2=16.49L
Answer:
Q = 28.9 kJ
Explanation:
Given that,
Mass of Aluminium, m = 460 g
Initial temperature, 
Final temperature, 
We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :
So, 28.9 kJ of heat is required to raise the temperature.
0.08 because 4mol NO divided by 50g gives you your answer
Answer:
9 (1-2x²)
Explanation:
The given expression is:
30 - 9x²*2 - 21 - 4 + 4
The first step is to compute the multiplication. This will give:
30 - 18x² - 21 - 4 + 4
Then, we will add like terms as follows:
(30-21-4+4) - 18x²
= 9 - 18x²
Finally, we can take the 9 as a common factor from both terms, this will give:
9 (1-2x²)
Hope this helps :)
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694
