Answer:
Explanation:C7H6O2 + (15/2) O2 = > 3H2O + 7CO2 delH = Σ stoichiometric coefficient* enthalpy of formation delH = (7*-393.509 kJ/mol) + (3*-285.83 kJ/mol) - (15/2 * 0) -(1*-385.2 kJ/mol) delH = -3226.853 kJ/mol benzoic acid ...
Answer:
9.10938356 × 10^-31 kilograms
Explanation:
i suck at chem lol
Answer:
Check the explanation
Explanation:
beam span = 25 ft
dead load = 0.6 kip/ft
live load = 2.1 kip/ft
factored load = 1.2*0.6 +1.6*2.1=4.08 kip/ft
moment in beam = 4.08*252/8=318.75 kip-ft = 3825 kip-in
design strength =0.9* 50 = 45 ksi
plastic section modulus required = 3825/45=85 in3
Moment in beam in ASD = (0.6+2.1)*252/8 = 210.9 kip-ft
lighest W section from LRFD = W21x44
lightest W section from ASD = W21x44
Answer:
R = 0.5825 k ohm
Explanation:
given data
vD = 0.75 V
iD = 1mA
resistor R = 15-V
solution
we get here first vD that is
vD =
vD = 0.825
so
iD = Ise ×
n = 1
so we can say
so it will
iD2 = iD1 ×
put here value and we get
iD2 = 1 ×
iD2 = 1 ×
iD2 = 20.086 mA
so
R will be
R =
R = 0.5825 k ohm
Attached is the solution to the question spelt out above.