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melamori03 [73]
2 years ago
10

Based on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the expe

rimental record. use complete sentences to explain the trend you predicted. you do not have to determine exact values for time; just describe the trend you would expect (increase or decrease) and why it occurs.
Chemistry
1 answer:
Anika [276]2 years ago
6 0

Factors that increases reaction rate such as increase in concentration or pressure will reduce reaction time whereas factors that decrease reaction rate such as inhibitors will increase reaction time.

<h3>What are the factors that affect reaction rate?</h3>

Factors that affect reaction rate are those factors which increase or decrease the rate of chemical reaction.

The factors that affect reaction rate include:

  • temperature
  • concentration/pressure
  • catalysts
  • surface area
  • nature of substance

Any factor that increases reaction rate such as increase in concentration or pressure will reduce reaction time whereas factors that decrease reaction rate such as inhibitors will increase reaction time.

Learn more about factors affecting reaction rate at: brainly.com/question/14817541

#SPJ1

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0.013M

Explanation:

m1v1/m2v2=n1/n2

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In which part of the cell is the majority of the energy released from the breakdown of glucose
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<em>Answer</em><em>:</em>

<em>Glycolysis</em>

<em>E</em><em>xplanation</em><em> </em><em>:</em>

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increase/increase

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3 years ago
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2ca(s)+o2(g) → 2cao(s) δh∘rxn= -1269.8 kj; δs∘rxn= -364.6 j/k
Alinara [238K]

Gibbs free energy change for the reaction at 29°C.  is equal to -1378.93 KJ.

<h3>What is Gibbs's free energy?</h3>

Gibbs free energy can be described as the enthalpy of the system minus the product of the temperature and entropy.

If the chemical reaction can be carried out under constant temperature ΔT = 0:

ΔG = ΔH – TΔS

The above equation is known as the Gibbs-Helmholtz equation.

ΔG > 0 non-spontaneous and endergonic and ΔG < 0 spontaneous and exergonic, ΔG = 0 is representing equilibrium.

Given the ΔS = -364 J/K, ΔH = -1269.8 KJ, T = 29°C = 29 + 273 = 302 K

ΔG = - 1269 - 302 × 364

ΔG =  -1269 KJ - 109.93 KJ

ΔG =  - 1378.93 KJ

Learn more about Gibbs's free energy, here:

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Your question is incomplete, most probably the complete question was,

2Ca(s)+O₂(g) → 2CaO(s)

ΔH∘rxn= -1269.8 kJ; ΔS∘rxn= -364.6 J/K

For this problem, assume that all reactants and products are in their standard states.

Calculate the free energy change for the reaction at 29°C.

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2 years ago
_____ NaPO4+_____KOH
Artist 52 [7]
NaPO4 + KOH -> KPO4 + NaOH
already balance
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