Answer:
a. 50ml b.10ml c. 6.097ml d. 190.1 ml
Explanation:
According to Boyle's law
Volume is inversely proportional to pressure at constant temerature
Mathematically
P1V1=P2V2
P1=Initial pressure=0.8atm
V1=Initial volume=25ml
making V2 the subject
at 0.4atm P2=0.4 atm,
V2=25×0.8/0.4
=50ml
at 2 atm V2=25×0.8/2
=10 ml
1mmHg=0.00131579
2500mmHg=3.28 atm
At 3.28 atm,V2=25×0.8/3.28
=6.097 ml
at 80.0 torr
1 torr=0.00131579
80 torr=0.1052 atm
at 0.1048 atm V2=25×0.8/0.1048
=190.1 ml
Answer:
bombarding it with an energetic particle
Explanation: nuclear reaction, a change in the identity or characteristics of an atomic nucleus, induced by bombarding it with an energetic particle. The bombarding particle may be an alpha particle, a gamma-ray photon, a neutron, a proton, or a heavy-ion.
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
An occluded front forms when a warm air mass is caught between two cooler air masses. The warm air mass is cut ofl or occluded' from the ground. The occluded warm front may cause clouds and precipitation. A swirling center of low air pressure is called a cyclone.
