B)Bacteria B)Bacteria B)Bacteria B)Bacteria
Answer:
591.471 MILLIMETERS
Explanation:
Formula
for an approximate result, multiply the volume value by 29.574
29.574*20=591.471
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Answer:
Heat of combustion =-7.35 *10² kJ
Explanation:
Step 1: Data given
Mass = 0.176 grams
Molar mass = 28.0 g/mol
Temperature increase 2.12 K
A bomb calorimeter has a heat capacity of 2.18 kJ/K
Step 2: Calculate heat of combustion
Energy released = mass *heat capacity* change in temperature
Q = c*∆T
Q = 2.18 kJ/K*2.12 K = 4.62 kJ released
Heat of combustion = (energy released)/(mole of substance)
Moles of gas = (0.176 g)/(28.0 g/mol) = 0.00629 moles
Heat of combustion = (energy released)/(mole of substance)
Heat of combustion = (4.62 kJ)/(0.00629 moles) = 7.35 *10² kJ
Heat of combustion =-7.35 *10² kJ (negative because heat is lost)
Answer:
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Explanation:
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