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:
a. The apparatus required to purify gypsum sample are: Bunsen burner, beaker, Filter Funnel, stirring rod, the filter paper.
b. Gypsum is a sulfate mineral that is made up of calcium sulfate dihydrate. Step-by-step instruction to purify gypsum sample is as follows:
1. Add water to the gypsum sample in a beaker.
2. Use the stirring rod to mix the mixture well.
3. Filter off the excess solid from the mixture using the filter paper and filter funnel.
4. Put the filtered mixture over the bunsen burner and evaporate the excess water from the mixture.
5. Allow the hot liquid to cool down and filter it again through the filter paper to get the pure gypsum.
A homogeneous mixture, the substances are uniformly distributed throughout the mixture
0.83 m/s seems the correct answer, hope it helps
Answer:
2.68 cm^3
Explanation:
Density= Mass/Volume
so...
8.96 g/cm^3 = 24.01 g/ V
and then u solve so it would be 2.68 cm ^3
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