Answer:
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Explanation:
Here, 
H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation
H(diene) depends on stability of diene.
More stable a diene, lesser will be it's H(diene) value (more neagtive).
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Hence, 
is higher (less negative) for trans-1,3-pentadiene
Put it in a beaker. Use a smaller beaker filled half way with ice and water and place in the larger one. It should be about an inch or two above the mixture. Heat over a Bunsen burner and the naphthalene will deposit on the bottom of smaller beaker.
And in this way, nephthalene be separated from the mixture of KBR and sand.
Answer:
1,085g of water
Explanation:
If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:
Q= m*C*Δt where;
Q in the heat, in this case: 4520kj
m is the mas
Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.
In order to determine the mass, I will have the same equation but finding m
m= Q/C*Δt without m=Q/C
So: m= 4,520J/4.18J/g°C
m= 1,0813 g
Given: C3H8(g) + O2(g) ----> CO2 (g) + H2O (g)
Step : Put a 3 in front of CO2 (g) to balance C
=> C3H8(g) + O2(g) ----> 3CO2 + H2O to balance H
Step 2: Put a 4 in front of H2O
=> C3H8 (g) + O2(g) -----> 3CO2 (g) + 4H2O (g)
Step 3: Given that there are 3*2 + 4 = 10 O to the right side, put a 5 in front of O2 to balance O:
=> C3H8(g) + 5O2(g) -----> 3CO2(g) + 4H2O(g)
You can verify that the equation is balanced.
So, the answer is that the coefficient in front of O2 is 5.
Losing eltron is the answer
