Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082
- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082 × 273 K
Solving:
n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
Answer:
Phenolphthalein is an indicator. It is pink in alkaline solutions and turns colourless as the pH decreases.
It can be used to measure the activity of the enzyme lipase on the breakdown of lipids.
Samples of milk containing phenolphthalein were incubated with lipase at different temperatures.
The time taken for the phenolphthalein to turn colourless was recorded and used to calculate the rate of enzyme activity.
Figure 10 shows these results.
Picture
(a) (i) Explain why phenolphthalein turns colourless when lipase breaks down the lipids in milk. (2)
(ii) Describe the effect of temperature on the activity of lipase, as shown in Figure 10. (2)
(iii) Explain why the activity of lipase changes above a temperature of 40°C. (2)
(b) A student investigated the time taken for amylase to breakdown a 10% starch solution into glucose at 37°C. The student repeated the investigation five times.
Answer:
5.4 tonnes.
Explanation:
The first step is to find the molar mass of Al2O3. Aluminum has a molar mass of about 27 and oxygen has a molar mass of about 16, so 2(27)+3(16)= 102g/mol=0.102kg/mol. 10200kg/0.102kg/mol=100,000 moles of Al2O3 in 10.2 tonnes. Multiplying this by the molar mass of the two aluminums, you get a total of 54*100,000=5400000g=5400kg=5.4 tonnes. Hope this helps!
Answer: option A. 2
Explanation: in the formula Sr3(PO4)2, the 2 behind (PO4) is affecting both P and O4. It means that we have P2 in the formula
