Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Answer:
i) 0.5071 (kg/s)
ii) -1407.1 kj/kg
iii) 204.05 Kw
iv) 5.881
v) 9.238
Explanation:
Given Data:
evaporation temperature ( T ) = 4°c = 277.15 K
Condensation Temperature ( T ) = 34°c = 307.15 K
<em>n</em> ( compressor efficiency ) = 0.76
refrigeration rate = 1200 kJ.s^-1
i) determine the circulation rate of the refrigerant
m = =
------- 1
Q = 1200 Kj.s^-1
H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )
H4 = entropy at step 4 = 142.4 ( kJ/ kg )
back to equation 1
m ( circulation rate of refrigerant ) = 0.5071 (kg/s)
ii) heat transfer rate in the condenser
Q = m ( H4 - H3 )
= 0.5071 ( 142.4 - 2911.27 )
= -1407.1 kj/kg
where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )
iii) power requirement
w = m * ΔH23
= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw
where: ΔH23 = =
= 402.37 (kj/kg)
iv) coefficient of performance of a cycle
W = Qc / w
= 1200 Kj.s^-1/ 204.05 kw
= 5.881
v) coefficient of performance of a Carnot refrigeration cycle
= 277.15 / ( 307.15 - 277.15 )
= 9.238
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change , we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
Thus, we obtain a quadratic equation whose solution is:
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.