The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
Mass of Product (y) vs Time (x) graph and Concentration of Reactant (y) vs Time (x) graph are two graphs which could be used to work out the rate of reaction from the gradient.
<h3>What is Rate of reaction ?</h3>
It is reffered as the speed at which a chemical reaction proceeds.
It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.
The rates of two or more reactions can be compared using a graph of mass or volume of product formed against time.
concentration of reactant against time of reaction is used to find the instantaneous rate of reaction at a given time.
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This is an incomplete question, here is a complete question.
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.
Answer : The work done during the combustion is, 9.9 kJ
Explanation :
Formula used :
where,
w = work done = ?
= change in enthalpy = -5074.2 kJ
= change in internal energy = -5084.1 kJ
R = gas constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
Thus, the work done during the combustion is, 9.9 kJ
There are several ways to prevent rusting. But you first need to know that rusting is caused by surrounding elements(wind,water,cold/heat etc)
So to prevent it, you need to have a coating protecting what you want to protect.
Some ways are to apply anti-rusting oil which is coating and another way I know that I know that they do for large buildings/structures such as bridges(the Golden gate bridge in San Fransisco is no exception) is to apply a special paint that protects then from rusting. But if there is even a little chip or crack in the paint, it can cause rusting in that area and slowly go under the paint and spread so workers have to sometimes repaint those spots to get rid of the rust.