Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

Answer:
1.
A= <u>sum</u><u>(</u><u>mass</u><u>*</u><u>percent</u><u> </u><u>abundance</u><u>)</u>
M 100
=(23.985*78.70)+(24.946*10.13)+(25.983*11.17)/100
= 24.3
2. The element is Magnesium.
3. 2412Mg,2512Mg and 2612Mg
<u>a) Answer: </u>
<em>Number of molecules in 1 mole</em>
<u>Explanation:</u>
a) Whether we take any of the substance among all three of the given substances they will have the same number of molecules in 1 mole of the substance is considered and the value for this will be 
<u>b) Answer: </u>
<em>In the given question </em><em>mass of the substance</em><em> which is </em><em>greatest</em><em> is asked for </em><em>one mole</em><em> and we also know that </em><em>mass of one mole is given by molar mass. </em>
<u>Explanation:</u>
b) It is known that
is the molar mass for oxygen which is greater than that of hydrogen while fluorine has a molar mass of
which on comparison shows that, it is the highest amongst all three.
<u>Answer:</u> The entropy change of the process is 
<u>Explanation:</u>
To calculate the entropy change for different phase at same temperature, we use the equation:

where,
= Entropy change
n = moles of acetone = 6.3 moles
= enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = ![-94.7^oC=[273-94.7]=178.3K](https://tex.z-dn.net/?f=-94.7%5EoC%3D%5B273-94.7%5D%3D178.3K)
Putting values in above equation, we get:

Hence, the entropy change of the process is 
Answer:
D.
Explanation:
-log(1.0x10^-5) = pH
pH + pOH = 14 (rearrange it)
OH- = 10^-pOH = 1.0 x 10^-9
- Hope that helped! Let me know if you need further explantion.