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.
Answer: B- 22.2 kg
Explanation: If three potatoes have mass of 667 g that means that each potato weighs 667/3= 222.33 g (approx) so 100 potatoes must be 100*222.33= 22233 g which equals 22.2 kg because 1 g=1000 kg
Answer:
Strong acids react faster where as weak acids take time to react with any base.
Answer:
See explanation
Explanation:
A titration involves the addition of a titrant to an analyte solution. It is a method of volumetric analysis.
When a particular volume of titrant is added, the colour changes to signal the end point of the reaction.
The point at which the colour changes is called the equivalence point. This is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.
Hence the volume NaOH that needs to be added to the beaker containing HCl to cause a colour change is the volume of NaOH that is just enough to completely neutralize the HCl solution.
