Which of the following statements about the atom is false?
2 answers:
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Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, 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.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082
- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082 × 273 K
Solving:
n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
Answer:
Check explanation
Explanation:.
NOTE: kindly check for attached file/picture for the graph.
From the graph of absorbance against concentration from the question. We can see that the 0.3 mark absorbance is equivalent to 0.15 M. So, the concentration of CuSO4 is 0.15 M.
The concentration can also be calculated using the Beer-lambert equation for absorbance. The equation is given below;
A= ɛ×C×l --------------------------------------------------------------------------------------(1).
Where A= absorbance, ɛ= molar absorptivity, C= concentration and l= length.
Therefore, the concentration,C will now be; C= A/ ɛ×l. -------------------------------------------------------------------------(2).
Assuming the length,l is 1cm.
Hence, C= 0.300/ ɛ×1.
C= (0.300/ ɛ) M.
