Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
The electrons in bonds (bonding domains) differ from lone
pairs (non-bonding domains) is because the bonding
domains are bonded to the central atom vs the lone pairs are just stuck on as
extra electrons. The difference of bonding domains from non-bonding
domains is that the bonding domains are bonded to the central atom and the
non-bonding domains are just stuck on as extra electrons.
Answer:
It takes 86 days take to cover half of the lake
Explanation:
In the day #1, the amount of the algae is X,
In the day #2 is 2X
In the day #3 is 2*2*X = X*2²
...
In the day #n the amount of the algae is X*2^(n-1)
Assuming X = 1m³. In the day 87, the area infected was:
1m³*2^(87-1)
7.74x10²⁵m³ is the total area of the lake
the half of this amount is 3.87x10²⁵m³
The time transcurred is:
3.87x10²⁵m³ = 1m³*2^(n-1)
Multiplying for 5 in each side:
ln (3.87x10²⁵) = ln (2^(n-1))
58.9175 = n-1 * 0.6931
85 = n-1
86 = n
<h3>It takes 86 days take to cover half of the lake</h3>
To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
The answer is B) heat will flow from the water to the copper until they both reach the same temperature
