Actual yield over theoretical yield, then multiply by 100
Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
<span>The correct answer is( A) blood.
when the buffer solution its PH value changes very little when a small amount
of strong acid or base is added to it, and here the bicarbonate buffering system is used to regular the PH of the blood that keeping the PH at nearly constant value by maintaining the original acidity or basicity of the solution.</span>
Answer:
Complete ionic: 
.
Net ionic: 
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water: 
, 
, and 
. These three salts will exist as ions:
- Each
formula unit will exist as one 
ion and one 
ion. - Each formula unit will exist as one
ion and two 
ions (note the subscript in the formula 
.) - Each formula unit will exist as one and two ions.
On the other hand, 
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite , , and (three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each formula unit will exist as only one ion and one ion. However, because the coefficient of 
in the original equation is two, 
alone should correspond to two 
ions and two 
ions.
Do not rewrite the salt because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of and two units of . Doing so will give:
.
Simplify the coefficients:
.
