Answer:-
As we can see from the graphical data,
The distance covered by all the four runners is the same 5 km.
Among the four athletes, Athlete P covers the distance in under three hours.
It is the minimum time taken among the four athletes.
Thus Athlete P covers the 5 km distance in the minimum amount of time.
We know that speed = 
Since time taken for P is minimum, his speed is the maximum. P ran the fastest.
Time taken by Q = 4.5 hours.
Speed of Q = 
= 
= 1.1 km/ hr
Time taken by R = 6 hours.
Speed of R = 
= 
= 0.8 km/ hr
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
<span>Releasing the oxygen from
the mercury does not change the mass. If you started with 20 g total
you would still have 20 g total after the heating</span>
In the given situation, the gas is heated under constant volume. As energy is supplied to the system in the form of heat, the frequency of collision between the gas particles increases. This increases the temperature of the gas consequently bringing about a decrease in pressure.
Based on the ideal gas law:
PV = nRT
Here, P/T = nR/V
If P1, T1 and P2, T2 are the pressure and temperature values before and after heating respectively, then since nR/V is a constant in this case, we have
P1/T1 = P2/T2 which is the Gay-Lussac's law.
