Given the reaction: S + O2 ---> SO2
1 answer:
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Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2
Taking into account about the ideal gas law, the mass of fluorine gas is 63.2244 grams.
<h3>What is ideal gas law</h3>An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature. The universal constant of ideal gases R has the same value for all gaseous substances.
- R is the ideal gas constant.
- n is the number of moles of the gas.
The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
<h3>Mass of fluorine gas</h3>In this case, you know:
- P= 850 mmHg= 1.11842 atm (being 1 mmHg= 0.00131579 atm)
- V= 45.5 L
- T =100 °C= 373 K (being 0°C= 273 K)
- R= 0.082
- n= ?
Replacing in the ideal gas law:
1.11842 atm ×45.5 L = n×0.082 × 373 K
Solving:
n= (1.11842 atm ×45.5 L)÷ (0.082 × 373 K)
<u><em>n= 1.6638 moles</em></u>
The molar mass of F₂ is 38 g/mole. Then you can apply the folowing rule of three: If by definition of molar mass 1 mole of the compound contains 38 grams, 1.6638 moles of the compound contains how much mass?
<u><em>mass= 63.2244 grams</em></u>
Finally, the mass of fluorine gas is 63.2244 grams.
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Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356