We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
<h3>
Answer:</h3>
0.34 mol S
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
11 g S
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.343 mol S ≈ 0.34 mol S
Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, 
The conjugate base is
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.

Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Answer:
0.0702J/g°C the specific heat capacity of the metal.
Explanation:m

where,
Q = heat absorbed by metal = 186.75 J
= Mass of metal= 19 g
= Initial temperature of metal = 
=Final temperature of metal = 
= specific heat of metal= ?



0.0702J/g°C the specific heat capacity of the metal.
Answer:
Option E!
Explanation:
If we were to draw the lewis dot structure for IBr2 -, we would first count the total number of valence electrons ( " available electrons " ). Iodine has 7 valence electrons, and so does Bromine, but as Bromine exists in 2, the total number of valence electrons would be demonstrated below;

Don't forget the negative on the Bromine!
Now go through the procedure below;
1 ) Place Iodine in the middle and draw single bonds to each of the bromine.
2 ) Add three lone pairs on each of the Bromine's
3 ) Now we have 6 electrons left, if we were to exclude the electrons shared in the " single bonds. " This can be placed as three lone pairs on Iodine ( central atom )!
The molecular geometry can't be linear, as there are lone pairs on the atoms. This makes it bent.