Answer:
C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)
Explanation:
Propane can be turned into hydrogen by the two-step reforming process.
In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:
C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)
In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:
CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)
In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.
C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)
C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)
Answer:
9.3 x 10^ -5 in standard notation is 0.000093
Explanation:
Since 10 is raised to a -5, this means that there should be 5 zeros before the non-zero digits
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
The entropy of the reaction can be calculated alike the enthalpy of the reaction which is equal to the difference between the summation of entropies of the products multiplied by their corresponding stoich coeff. and the summation of the entropies of the reactants multiplied by their corresponding stoich coeff. In this case, the answer is -146.8 J / mol K
