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n200080 [17]
2 years ago
5

Calculate the molar mass of menthol, C10H20O

Chemistry
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

C10H200

Explanation:

The molecular formula C10H20O (molar mass : 156.27 g/mol) may refer to: Citronellol. Decanal.

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Suggest the pH of the solution made when potassium oxide dissolves in water.
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Since KOH is a strong base, the solution completely ionizes into K+ and OH- when in water. The reaction KOH --> K+ + OH- takes place. The concentration of [ OH- ] can then be used to calculate the pOH of the solution. pH = 14 - pOH = 14 - 1.48 = 12.52

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101. During the making of steel, iron(II) oxide is reduced to metallic iron by treatment with carbon monoxide gas.
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Please comment the correct answer and I'll give you brainliest! Which atom of each pair below has the largest ionization energy?
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Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
3 years ago
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