The heat released by the gas is -225 J
Explanation:
First of all, we have to calculate the change in internal energy of the gas, which for a diatomic gas is given by

where
n = 0.235 mol is the number of moles
is the gas constant
is the change in temperature
Substituting,

Now we can us the 1st law of thermodynamics to find the heat absorbed/released by the gas:

where
is the change in internal energy
Q is the heat
W = 205 J is the heat done by the gas
Solving for Q,

Since the sign is negative, it means the heat has been released by the gas.
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Answer:
N≡N bond
Explanation:
Bond energy (bond enthalpy) is a measure of the bond strength in the bond. It is defined as average value of bond dissociation energies in the gas-phase for all the bonds having same type with in same chemical species.
<u>The greater the bond energy, the greater is the amount of energy required to break the bond, the more stable is the bond.</u>
Thus, among the following bonds:
O=O 498 kJ/mol
N≡N 946 kJ/mol
C=C 614 kJ/mol
C=O 745 kJ/mol
C≡C 839 kJ/mol
<u>The greatest bond energy is of N≡N 946 kJ/mol and thats why it is the most stable.</u>
Low pressure has a bit less of a function than high pressure, high pressure is more useful in certain terms
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole

- P is power
- R is resistance

Hence


- Therefore if power is low then resistance will be high.
The first bulb has less power hence it has greater filament resistance.