Given question is incomplete. The complete question is as follows.
Pentaborane (
) is a colorless highly reactive liquid that will burst into flames when exposed to oxygen.the reaction is:
Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.the standard enthalpy of formation 
, 
, and 
are 73.2, -1271.94, and -285.83 kJ/mol, respectively.
Explanation:
As the given reaction is as follows.
Therefore, formula to calculate the heat energy released is as follows.
Hence, putting the given values into the above formula is as follows.
= 
= -9078.59 kJ/mol
Since, 2 moles of Pentaborane reacts with oxygen. Therefore, heat of reaction for 2 moles of Pentaborane is calculated as follows.
= -71.915 kJ/g
Thus, we can conclude that heat released per gram of the compound reacted with oxygen is 71.915 kJ/g.
Wave A has the shorter wavelength due too the fact that the shorter the wavelength the higher the frequency as they have an inverse relationship.
Answer:
37
Cl
17
(37 is a superscript to the left of the symbol, and 17 is a subscript to the left of the symbol)
Explanation:
1) Chlorine-37 is the chlorine isotope with mass number 37.
2) The atomic mass of chlorine is 17. All the atoms of chlorine have the same atomic number (all the atoms of an element have the same atomic number). You find this number in a periodic table.
3) The isotope representation consists of:
i) The symbol of the element (Cl in this case) at the center of the representation.
ii) The mass number (37) as a superscript to the left of the symbol. That is:
37
Cl
iii) The atomic number (17) as a subscript to the left of the symbol. That completes the representation in this form:
37
Cl
17
Answer:
PCl5 <-> PCl3 + Cl2
PCl5 decomposes into PCl3 and Cl2 according to the equation above. A pure sample of Pcl5 is placed in a rigid, evacuated 1.00 L container. The initial pressure of the PCl5 is 1.00 atm. The temperature is held constant until the PCl5 reaches equilibrium with its decomposition products. The figures below show the initial and equilibrium conditions of the system.
As the reaction progresses toward equilibrium, the rate of the forward reaction
A) increases until it becomes the same as the reverse reaction rate at equilibrium
B) stays constant before and after equilibrium is reached
C) decreases to become a constant nonzero rate at equilibrium
D) decreases to become zero at equilibrium
Explanation:
At equilibrium, both forward and backward reactions take place with constant speed.
The reaction will never cease.
Due to this reason chemical equilibrium is called dynamic in nature.
At equilibrium:
the rate of forward reaction = rate of backward reaction
As the reaction progresses toward equilibrium, the rate of the forward reaction decreases to become a constant nonzero rate at equilibrium.
Answer is option C).
It's called the <span>Filament.</span>
