Answer:
ΔH = - 5315 kJ.
Explanation:
The given chemical reaction is as follows -
2C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g) + 5315 kJ
In the above equation , the amount of energy i.e. 5315 kJ is released , i.e. it is in the product side , hence , the reaction is an example of an exothermic reaction .
Hence ,
The value of the change in enthalphy , i.e. , the enthalpy of product minus the enthalpy of the product .
Therefore ,
The value of the change in enthalphy = - ve .
Hence ,
ΔH = - 5315 kJ.
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
- Heat absorbed (Q): 53.1 J
- External pressure (P): 0.677 atm
- Final volume (V2): 63.2 L
- Change in the internal energy (ΔU): -108.3 J
Step 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
<span>let x=gallons of current mixture to be drained
and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.</span>
<span>
0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal
Thus, 1.65 gallons of current mixture to be drained and replaced with pure
antifreeze.</span>
Because the chemicals are different
