Answers:
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Answer:
49.4 g Solution
Explanation:
There is some info missing. I think this is the original question.
<em>A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g available of a 40.5 % w/w solution of acetic acid in acetone. </em>
<em>
Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.</em>
<em />
We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:

Okay
Mr (H2O)= 18g
therefore moles of H2O
is 720.8/18= 40.04mol
the ratio of H2 to O2 to H2O is
2 : 1 : 2
so moles of H2 is same as H2O here
H2= 40.04moles
moles of O2 is half
so 40.04 x 0.5
20.02moles
grams of O2 is
its moles into Mr of O2
that's 20.02 x 32 = 640.64g
<u>Answer:</u> The
of the reaction at given temperature is -12.964 kJ/mol.
<u>Explanation:</u>
For the given chemical reaction:

The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 
Putting values in above equation, we get:

Hence, the
of the reaction at given temperature is -12.964 kJ/mol.
Answer:
Accuracy
Explanation:
It is important for scientists to review the work of other scientists, so they can be sure there are no mistakes or lack of judgement. They repeat to compare results to make hypotheses.