The result of multiplying that equation is 70,700.
It is the 4th one that is correct and balanced
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.
It would be +7 oxidation number.
[All compounds are zero]
Al is in group III [ or 13 depending on the system ] so the oxidation state is 3.
You have 12 O at -2 each for-24. That leaves you at -24+3=-21. So, 3 Cl atoms must be +21 to balance the -21 there which makes +21/3 or +7 for each Cl.
Answer:
A. change speed, direction, or both
Explanation:
hope this helps. . . <3
good luck! uwu
