In a laboratory, you scratch an unknown mineral on a piece of unglazed porcelain. Which property are you trying to determine?
2 answers:
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Considering the ideal gas law and STP conditions, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.
<h3>Definition of STP condition</h3>The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<h3>Ideal gas law</h3>Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
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.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
In first place, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide, 5.13 L (5.13 dm³= 5.13 L, being 1 dm³= 1 L) are occupied by how many moles of carbon dioxide?
<u><em>amount of moles of carbon dioxide= 0.229 moles</em></u>
Then, you know:
- P= 286.5 kPa= 2.8275352 atm (being 1 kPa= 0.00986923 atm)
- V= ?
- T= 12.9 C= 285.9 K (being 0°C= 273 K)
- R= 0.082
- n= 0.229 moles
Replacing in the ideal gas law:
2.8275352 atm× V = 0.229 moles×0.082 × 285.9 K
Solving:
V= (0.229 moles×0.082 × 285.9 K)÷ 2.8275352 atm
<u><em>V= 1.8987 L</em></u>
Finally, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.
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STP conditions:
the ideal gas law:
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Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is =
Now after adding extra 10 mL , the density becomes .
Therefore,
So the density decreases when we add more solution.