C3H8.gas reacts with 5L of O2 at STP
Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.
Explanation:
The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.
Answer:

Explanation:
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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

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Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Answer:
Lateral
Explanation:
Is the opposite side of the midline