Answer:
The volume is 4.13793 L
Explanation:
Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:
Density is a characteristic property of every body or substance.
The most commonly used units of density are 
or 
for solids, and 
or 
for liquids and gases.
In this case, you know:
- density= 0.87
- mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)
- volume= ?
Replacing:
Solving:
volume= 4,137.93 mL
Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L
<u><em>The volume is 4.13793 L</em></u>
Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
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
I love these type of questions :)
Final Answer: Neon
