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3 years ago

Which provides a baseline for an experimental investigation

Chemistry

1 answer:

Goryan [66]3 years ago

7 0

Answer:

control group

Explanation:

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points

labwork [276]

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

$Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

$NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

$H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

$NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$