The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O
M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5
moles of HCl = 5.5/37.5=0.14
M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)
M(C2H6O)= 2x12+ 6 + 16=46,
moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles
</span>
the sum of the moles is 0.14 + <span>4.35 </span> = 4.501 moles
finally, <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
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because it can be found by 0.14 / 4.501= 0.031
Answer:
The reaction will be spontaneous
Explanation:
To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:
<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>
Calculating the
:
Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.
In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in:
If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.
Explanation:
Plasma is an ionized gas, a distinct fourth state of matter. “Ionized” means that at least one electron is not bound to an atom or molecule, converting the atoms or molecules into positively charged ions.but plasma is present in human body too in a liquid form.
hope this helps you
have a nice day
Answer:
the value of H° is below -6535 kj. +6H2O
Explanation:
6H2O answer solved
Answer:

Explanation:
Hello,
In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

Best regards.