Flerovium at its ground state is solid. It has electron configuration of [Rn]5f¹⁴6d¹⁰7s²7p². The expected number of valence electrons in a flerovium atom is 2. A ground state is the most stable state of an atom at satndard temperature and pressure.
<u>Answer:</u> The
for the reaction is 51.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of carbon and water follows:

The intermediate balanced chemical reaction are:
(1)
( × 2)
(2)
( × 2)
(3)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B2%5Ctimes%20%5CDelta%20H_1%5D%2B%5B2%5Ctimes%20%5CDelta%20H_2%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-393.7%29%29%2B%282%5Ctimes%20%28-285.9%29%29%2B%281%5Ctimes%20-%28-1411%29%29%5D%3D51.8kJ)
Hence, the
for the reaction is 51.8 kJ.
Answer:
colloidal system liquid dispersed in gas.
Answer:
solid
Explanation:
Melting and boiling points of Group 7 elements State at room temperature Room temperature is usually taken as being 25°C. At this temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine and astatine are solids. There is therefore a trend in state from gas to liquid to solid as you go down the group.
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!