Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two. Where: ¯v = average velocity.
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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.
1. Atoms are mostly empty space
2. Atoms have a concentration of positive matter somewhere in them
3. The concentration of this positive matter is very small
Answer:
we use atom in our sentences all the time
Explanation:
that what i put
