Answer:
See the answer below, please.
Explanation:
The equilibrium constant is defined as the relationship between products and reagents, each one elevated to their stoichiometric coefficients, in that of the given equation, the Kc is:
Kc= (NH4)^1/ (NH3)^1 x (HI)^1
NH4= products
NH3 and HI = reagents
Freidal craft reaction is the attack of a carbon or carbon chain on aromatic ring with the help of anhydrous AlCl3 to produce alkylated benzene ring.
Only ketone not be able to undergo friedal craft reaction as, it is not aromatic compound whereas all the given reactants are aromatic and gives friedal craft reaction.
Answer:
B) an anion
Explanation:
During bonding atoms may share electron or give out electron . In the case of electrovalent or ionic bonding, electron(s) is/are transferred from one atom to another . The atom that donate the electron are known as donor atom and the atom that receives the electron are known as receivers.
The atom that donate the electron becomes a cation(positively charged) as it lose it electron(s) while the receiver gains electron to become an anion which is a negatively charge ion.
An example of such interaction is between a sodium atom and a chlorine atom. The Na atom lose it electron to become positively charged(cation) while the chlorine atom gains electron to become negatively charged(anion). Both atom bond together losing and gaining electron to fulfill the octet rule.
Na+ + Cl- → NaCl
Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
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I hope it helps!