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3 years ago

A substance is followed by the symbol (I) in a chemical Equation. What does the symbol Represent

Chemistry

1 answer:

Ber [7]3 years ago

8 0

Answer:

Iodine

Explanation:

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The common pathway for the oxidation of glucose and fatty acids is ________.

finlep [7]

The citric acid cycle should be the answer.

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3 years ago

Fill in the blank.

Karolina [17]

Answer:

Number of valence electrons

Explanation:

A family of elements is defined as a specific group of those elements. Groups in the periodic table are identified by columns, e. g., group 1A contains alkali metals, group 2A contains alkaline earth metals, group 7A consists of halogens, group 8A contains noble gases etc.

The group number corresponds to the valence electrons these atoms have. The number of protons for each atom is different and the level of reactivity depends on the size of the atom mainly, as well as electronegativity.

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4 years ago

Read 2 more answers

Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need

Assoli18 [71]

Answer:

$\Delta H_{f}^{\circ } (HNO_{3}, aq)$ = (-207.39) kJ/mol

Explanation:

Given:

$\Delta H_{f}^{\circ } (NH_{3}, g)= -46.1 kJ/mol\\ \Delta H_{f}^{\circ } (H_{2}O, l)= -285.8 kJ/mol\\ \Delta H_{f}^{\circ } (HNO_{3}, aq)= ? kJ/mol$

3NO₂(g)+H₂O(l)→ 2HNO₃(aq)+NO(g); ΔH° = -137.3 kJ ....equation 1

2NO(g)+O₂(g)→ 2NO₂(g) ; ΔH° = -116.2 kJ ....equation 2

4NH₃(g)+5O₂(g)→ 4NO(g)+6H₂O(l) ;ΔH° = -1165.2 kJ ....equation 3

Multiplying equation 1 with (2/3), we get

2NO₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) + (2/3) NO(g) ....equation 4

Adding equation 2 and 4, we get

2NO₂(g) + 2/3 H₂O(l) + 2NO(g) + O₂(g) → 4/3 HNO₃(aq) + (2/3) NO(g) + 2NO₂(g)

⇒ (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) ....equation 5

Multiplying equation 3 with (1/3), we get

(4/3) NH₃(g) + (5/3) O₂(g) → (4/3) NO(g)+ 2 H₂O(l) ....equation 6

Now adding equation 5 and 6, we get

(4/3) NH₃(g) + (5/3) O₂(g) + (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → (4/3) NO(g)+ 2 H₂O(l) + 4/3 HNO₃(aq)

⇒ (4/3) NH₃(g) + (8/3) O₂(g) → (4/3) HNO₃(aq) + (4/3) H₂O(l) ....equation 7

Now multiplying equation 7 with (3/4), we get

NH₃(g) + 2 O₂(g) → HNO₃(aq) + H₂O(l) ...equation 8

Therefore, by Hess's law the standard enthalpy of formation is:

ΔH° = (3/4) [ (-137.3 kJ) × (2/3) + (-116.2 kJ) + (-1165.2 kJ) × (1/3)]

ΔH° = (3/4) [ - 91.53 - 116.2 - 388.4]

ΔH° = (3/4) [-596.13] = -447.09 kJ

Since the change in enthalpy of a reaction:

ΔH° = $[\Delta H_{f}^{\circ } (HNO_{3}, aq) + \Delta H_{f}^{\circ } (H_{2}O, l)] - [\Delta H_{f}^{\circ } (NH_{3}, g) + 2 \Delta H_{f}^{\circ } (O_{2}, g)]$