Answer:
Mass = 0.32 g
Explanation:
Given data:
Mass of CH₄ = ?
Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)
Temperature = 273 K
Pressure = 1 atm
Solution:
Volume of CH₄:
500 mL (500 mL× 1L/1000 mL= 0.5 L)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K
0.5 atm.L = n×22.4 atm.L/ mol
n = 0.5 atm.L / 22.4 atm.L/ mol
n = 0.02 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.02 mol × 16 g/mol
Mass = 0.32 g
1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa
101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm
1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg
Answer:
From the graph find the maximum velocity and half it i.e. Vmax/2. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point. This will give the value of Km.
Answer: 3 <span>moles of water would be produced in present case.
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Reason:
Reaction involved in present case is:
<span> C5H12 + 8O2 </span>→<span> 5CO2 + 6H2O
In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.
Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.</span>
