Given the reactions, x(s)+12o2(g)⟶xo(s) δ=−549.3 kj x(s)+12o2(g)⟶xo(s) δh=−549.3 kj xco3(s)⟶xo(s)+co2(g)δ=+170.9 kj xco3(s)⟶xo(s)+co2(g)δh=+170.9 kj what is δδh for this reaction? x(s)+12o2(g)+co2(g)⟶xco3(s)

-720.2 kJ

Explanation:

We have two equations:

(I)   X + ½O₂ ⟶ XO;       ΔH = −549.3 kJ

(II) XCO₃ ⟶ XO + CO₂; ΔH =  +170.9 kJ  

From these, we must devise the target equation:

(III) X +½O₂ +CO₂ ⟶ XCO₃; ΔH = ?

The target equation has X on the left, so you rewrite Equation(I).

(I) X + ½O₂ ⟶ XO; ΔH = −549.3 kJ

Equation (IV) has XO on the right, and that is not in the target equation.

You need an equation with XO on the left, so you reverse Equation (II).  

When you reverse an equation, you reverse the sign of its ΔH.  

(IV)  XO + CO₂ ⟶ XCO₃ ;  ΔH = -170.9 kJ  

Now, you add equations (I) and (IV), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation (III):

(I)   X + ½O₂ ⟶ XO;                ΔH = -549.3 kJ

(IV) XO + CO₂ ⟶ XCO₃ ;       ΔH =  -170.9 kJ  

(III) X + ½O₂ + CO₂ ⟶ XCO₃; ΔH = -720.2 kJ

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

the answer is c

tell me if wrong