Answer:
Large temperature and air pressure decrease.
Temperature and air pressure increase.
Explanation:
Answer:
The new pressure of the pump is 26.05 atm or 2639.4 kPa
Explanation:
Step 1: Data given
Volume of the bicycle tire pump = 252 mL = 0.252 L
Pressure of air = 995 kPa = 9.81989 atm
The volume of the pump is reduced to 95.0 mL = 0.095 L
Step 2: Calculate the new pressure
V1*P1 = V2*P2
⇒with V1 = the initial volume of the bicycle tire pump = 0.252 L
⇒with P1 = the initial pressure of the pump = 9.81989 atm = 995 kPa
⇒with V2 = the reduced volume of the pump = 0.095 L
⇒with P2 = the new pressure = TO BE DETERMINED
0.252 L * 9.81989 atm = 0.095 L * P2
P2 = 26.05 atm
The new pressure is 26.05 atm
OR
0.252 L * 995 = 0.095 L * P2
P2 = 2639.4 kPa
The new pressure of the pump is 26.05 atm or 2639.4 kPa
About 6.5x10^22 molecules.
(5g C2H5OH)x(1 mol C2H5OH/46g C2H5OH)x(6.02x10^23 molecules C2H5OH/1 mol C2H5OH)=(3.01E24)/46=6.5x10^22.
Let me know if this helped!
Which fractions are equivalent to 0.875?
