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
?? Is that the whole question?
x = 1.01
Explanation:
Given equation:
y = 1.2345x – 0.6789
y = 0.570
Problem:
Solving for x
The variables in this equation are y and x
They can take up any value since they are variables.
Since we have been given y = 0.570
y = 1.2345x – 0.6789
To solve for x, we simply substitute for y in the equation;
since y = 0.570
0.57 = 1.2345x – 0.6789
add 0.6789 to both sides;
0.57 + 0.6789 = 1.2345x – 0.6789 + 0.6789
1.2489 = 1.2345x
Divide both sides by 1.2345
x = 1.01
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An isotope has the same number of <em>protons</em> but a different number of <em>elec</em><em>tro</em><em>n</em><em>s</em><em> </em>than others atom of the same element