You are breathing because you are tired from running.
Answer:
There are 9.8 ×1021. formula units in 10,005.8gCaO
A substance that produces hydrogen ions is an acid
Answer:
Products are favored.
Explanation:
The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = ?
It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵
NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹
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CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = 1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>
<em> </em>
As Kr is defined as:
Kr = [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]
And Kr is > 1
[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],
showing <em>products are favored</em>.
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
