Answer:
I think it's the second answer --If you increase the acidity..
I hope answer I can answer your question!
Answer:

Explanation:
Given: Entropy of surrounding: ΔSsurr = ?
Temperature: T= 355 K
The change in enthalpy of reaction: ΔH = -114 kJ
Pressure: P = constant
As we know, ΔH = -114 kJ ⇒ negative
Therefore, the given reaction is an exothermic reaction
Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,


<u><em>In the given reaction:</em></u>
2NO(g) + O₂(g) → 2NO₂(g)
As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.
As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>
<u>Therefore, at the given temperature,</u>
- if
then the given reaction is spontaneous
- if
then the given reaction is non-spontaneous
Answer:
when atoms lose or gain electrons
Explanation:
Answer:
Therefore, the boiling points of the alkanes increase with molecular size. For isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors.
The boiling points of the normal alkanes increase with increasing molecular weight (Table 3.3). As the molecular weight increases, London forces increase because more atoms are present to increase the surface area or the molecules.
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Answer:
<em>2.54 mol CO2 </em>
Explanation:
To calculate the mol of carbon dioxide that will be generated by megajoule (MJ). First, we have to find the value of the formation enthalpy for the CO2 at standard conditions.
<em>Remember that standard conditions are at 25°C and 1 atm. </em>
Any physic chemistry book has the table of ΔHf°, I always use one name Levine's physic chemistry. You can trust its data.
Enthalpy of formation for CO2: -393.509 kJ/mol
As you can see this value means that for each 393.509 kJ liberate (that's indicated for the menus sing), 1 mol of CO2 is formed.
Let's write this using mathematic language:
393.509 kJ/ 1 mol CO2
The problem asks you to find how many mols are produced by 1 megajoule liberate, let's change megajoule to kilojoule.
1 megajoule (MJ) = 1000 kJ
1000 kJ x (1mol/ 393.509 kJ) = 2.54 mol CO2
<em>Remember don't use the negative value to make the calculus. The negative value just indicates that the heat is been released.</em>
If you want to watch the mathematics details, go to the image attached.