Answer:
C. 30 kJ
Explanation:
Hello there!
In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

Therefore, the answer is C. 30 kJ
.
Best regards!
Explanation:
The given data is as follows.
P = 3 atm
=
= 
= 9 L =
(as 1 L = 0.001
),
= 15 L = 
Heat energy = 800 J
As relation between work, pressure and change in volume is as follows.
W = 
or, W = 
Therefore, putting the given values into the above formula as follows.
W = 
= 
= 1823.85 Nm
or, = 1823.85 J
As internal energy of the gas
is as follows.
= Q - W
= 800 J - 1823.85 J
= -1023.85 J
Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:
A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4
As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.
Answer:
See explaination and attachment
Explanation:
please kindly see attachment for the step by step solution of the given problem.
Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: <u>Bubbles</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>colourless</u><u> </u><u>gas</u><u> </u><u>(</u><u>carbon</u><u> </u><u>dioxide</u><u> </u><u>gas</u><u>)</u>
(ii) CH3CH2CH2OH and KMnO4 /H
observation: <u>The</u><u> </u><u>orange</u><u> </u><u>solution</u><u> </u><u>turns</u><u> </u><u>green</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>oxidation</em><em> </em><em>of</em><em> </em><em>propanol</em><em> </em><em>to</em><em> </em><em>propanoic</em><em> </em><em>acid</em><em> </em><em>occurs</em>]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: <u>A</u><u> </u><u>sweet</u><u> </u><u>fruity</u><u> </u><u>smell</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: <u>a</u><u> </u><u>brown</u><u> </u><u>solution</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>