Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
Learn more: brainly.com/question/2510654
Answer:
OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20
Explanation:
Butane burns is air (O2) according to the equation:
C4H10 + O2 —> CO2 + H20
Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;
There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:
C4H10 + O2 —> 4CO2 + H20
Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:
C4H10 + O2 —> 4CO2 + 5H20
Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2
as shown below
C4H10 + 13/2O2 —> 4CO2 + 5H20
Now we multiply through by 2 clear off the fraction and we obtained:
2C4H10 + 13O2 —> 8CO2 + 10H20
Answer:
Final temperature of the gas = -146.63 °C
Explanation:
At constant pressure, volume and temperature of the gases are related as:
Where,
V1 = Initial volume = 1.00 L
V2 = Final volume = 2.40 L
T1 = Initial temperature = 30.5 °C = 30.5 + 273.15 = 303.65 K
Now, substitute the values in the above equation,
T2 = 126.52 K
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
T( °C) = T(K) - 273.15
= 126.52 - 273.15 = -146.63 °C
