Answer:
3 moles of calcium oxide are formed
Explanation:
Based on the reaction of Calcium Ca, with oxygen, O₂, to produce calcium oxide, CaO is:
2Ca + O₂ → 2CaO
<em>where 2 moles of calcium produce 2 moles of Calcium oxide.</em>
That means, if 3 moles of Ca react in presence of oxygen:
3 moles of calcium oxide are formed
If an Atom has 12 Electrons then if it is a Neutral atom then the Atomic Number of the Atom will be 12.
The Electronic Configuration of Element with Atomic Number 12 will be :
⇒ 1s² 2s² 2p⁶ 3s²
We can notice from the Electronic Configuration that :
Given atom has 2 electrons in the 1st shell.
⇒ In the Shell : n = 1
Number of Electrons = 2
In the similar way : Number of Electrons in 2nd shell = 2 + 6 = 8
⇒ In the Shell : n = 2
Number of Electrons = 8
In the similar way : Number of Electrons in 3rd shell = 2
⇒ In the Shell : n = 3
Number of Electrons = 2
Surely, If cells did not work together in an organism, there won't be formation of new cells and life process would stop
<h3>Living organisms </h3>
Living organisms; be it plants or animals are any organic or living system composed of cells and function as an individual entity.
- All living organisms share a number of key characteristics or functions such as movement, respiration, homeostasis, reproduction, growth, evolution, competition and others.
- Animals and plants also posess systems such as the digestive, skeletal, transport, nervous, excretory, respiratory and reproductive system.
- Living organisms are also taxonomically classified as either unicellular microorganisms or multicellular plants and animals
So therefore, surely, If cells did not work together in an organism, there won't be formation of new cells and life process would stop
Learn more about living organisms:
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<u>Answer:</u> The amount of energy released per gram of 
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of produces -9078.57 kJ of energy.
Or,
If 
of produces -9078.57 kJ of energy
Then, 1 gram of will produce = 
of energy.
Hence, the amount of energy released per gram of is -71.92 kJ
Halogen--Florine
Chalogen-- Oxygen
Alkali Metal-- Sodium
Boron -- Metalloid (atomic symbol B)
Lanthanide series-- (Number 57-71 on periodic Table) Example: Cerium #58
Alkaline Earth Metal--Magnesium
Transition Metal-- Gold, Iron, Silver. Etc... (Middle section of a period table)
Nobel Gas--Helium
