Answer:
1 mole of C2H6.
Explanation:
The balanced equation for the reaction is given below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
We can determine the number of mole of C2H6 that reacted to produce 2 moles of CO2 as follow:
From the balanced equation above,
2 moles of C2H6 reacted to produce 4 moles of CO2.
Therefore, Xmol of C2H6 will react to produce 2 moles of CO2 i.e
Xmol of CO2 = (2 x 2)/4
Xmol of CO2 = 1 mole.
Therefore, 1 mole of C2H6 is required to produce 2 moles of CO2.
B. Heating up the reaction will increase the entropy of a reaction.
<h3>
What is entropy?</h3>
Entropy is the measure of the degree of disorderliness of a system.
Entropy is also the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
S = ΔH/T
where;
- S is entropy
- ΔH is energy input
- T is temperature
Entropy increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules.
However, entropy increases as temperature increases. Thus, heating up the reaction will increase the entropy of a reaction.
Learn more about entropy here: brainly.com/question/6364271
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Altitude. Rainfall, a hot dry day, and a hurricane are all examples of weather, not climate.
i think its MIDDLE FINGERS UP IN THE SKY AND AT THESE AHOLE MODERATORS
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol
HNO3= 1 mol KOH (keep in mind this because it will be used later).
We also know that 0.100 M KOH aqueous
solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the
definition of molarity).
First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/
1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.
Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.
The final answer is </span>(2) 20.0 mL.<span>
Also, this problem can also be done by using
dimensional analysis.
Hope this would help~
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