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Colt1911 [192]
3 years ago
15

For the reaction A+B↽−−⇀C+DA+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing the condit

ions of the reaction can alter the value of the change in free energy (Δ????)(ΔG) . Classify the conditions as to whether each would decrease the value of Δ????ΔG , increase the value of Δ????ΔG , or not change the value of Δ????ΔG for the reaction. For each change, assume that the other variables are kept constant.
Chemistry
2 answers:
BaLLatris [955]3 years ago
7 0

Explanation:

Gibbs free energy (G) is dependent upon the enthalpy (H) , temperature (T) and entropy (S) as shown in the equation below;

ΔG =  ΔH   −  TΔS

The factors affect ΔG of a reaction are given below;

Decrease the Value of G

- Positive ΔH and Positive ΔS at high temperatures.

- Negative ΔH and Negative ΔS at low temperatures.

Would not change the value of G

- Negative values of ΔH and Positive values of ΔS; (at all temperatures,  ΔG would be Negative)

- Positive values of ΔH and Negative values of ΔS; (at all temperatures,  ΔG would be positive)

Increase the Value of G

- Positive values of ΔH and Positive ΔS at low temperatures.

- Negative ΔH and Negative ΔS at high temperatures.

77julia77 [94]3 years ago
5 0
What are the answer choices
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Predict the products of each of these reactions and write balanced complete ionic and net ionic equations for each. If no reacti
Bumek [7]

Answer:

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq)  →  AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq)  + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq)  →  AgCl(s) + Li⁻(aq)  + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) →  AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq)  →  Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq)  + SO²⁻₄(aq) + 2Li⁺(aq)  + SO₃²⁻(aq)  →  2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq)  →  SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq)  →  Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq)  →  2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq)  →   LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq)  →   3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq)  →   3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) +  3OH⁻ (aq)  →    Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq)  → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq)   → Hg₂Cl₂ (s)

8 0
3 years ago
If an atom has 34 protons and 40 neutrons, what is its mass number? 40 6 74 34
Darya [45]
Protons+Neutrons
34+40= 74
7 0
3 years ago
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A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but does not put fertilizer on row 2. Both
IrinaVladis [17]

Answer:

Q9. The independent variable in this experiment is the fertilizer. It is independent because she manipulating the variable to compare the growth.

Q10. The dependent variable in this experiment is the amount of growth of the corn. It is this because the growth depends on what the scientist did on the corn.

Q11. The variable controlled in this experiment is the amount of sun and water. These two variables never change so this is why it is the control.

Explanation:

5 0
3 years ago
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Electrons involved in bonding between atoms are____.
mariarad [96]

Answer:

valence electrons

8 0
3 years ago
You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0
3 years ago
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