Distance = 556 km
Time = 3.4 h
Speed = Distance / Time = 556 / 3.4 = 163.52 km/h
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
1. The reaction is exothermic.
2. I know this because the delta H is a negative number. (-802.4 kJ)
3. This means that this reaction releases heat.
The lack of a number preceding the carbon symbol C and the compound formula CO2 shows that there is one carbon atom and one carbon dioxide molecule. Subscript numbers in chemical formulas represent the number of atoms or molecules immediately preceding the subscript.
<u>Answer:</u>
<u>For 2:</u> The % yield of the product is 92.34 %
<u>For 3:</u> 12.208 L of carbon dioxide will be formed.
<u>Explanation:</u>
The percent yield of a reaction is calculated by using an equation:
......(1)
Given values:
Actual value of the product = 78.4 g
Theoretical value of the product = 84.9 g
Plugging values in equation 1:
Hence, the % yield of the product is 92.34 %
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(2)
Given mass of carbon dioxide = 24 g
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:
<u>At STP conditions:</u>
1 mole of a gas occupies 22.4 L of volume
So, 0.545 moles of carbon dioxide will occupy = of volume
Hence, 12.208 L of carbon dioxide will be formed.