Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Idk but look it up on google
Hello there!
The statement that Acid strength in a series of H-A molecules increases with increasing size of A is True.
When only the size is involved, increasing the size will increase the Acid strength because as size increases, the H-A bond will become weaker as the atoms will be farther apart. Acid strength is related to the ability to release H⁺ ions and a weaker H-A bond will release H⁺ more easily.
Have a nice day!
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.
