Answer:
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Burning paper , shredding plastic , burning a match
Answer:
56972.17K
Explanation:
P = 4.06kPa = 4.06×10³Pa
V = 14L
n = 0.12 moles
R = 8.314J/Mol.K
T = ?
We need ideal gas equation to solve this question
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
T = PV / nR
T = (4.06×10³ × 14) / (0.12 × 8.314)
T = 56840 / 0.99768
T = 56972.17K
Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa
Answer:
- The annual cost for electricity to power a lampost for 7.00 h per day with a 100 W incandescent bulb is $ 40.89.
- The annual cost for electricity to power the same lampost for the same time with an energy efficient 25 W fluorescent bulb is $ 10.22.
So, you can save $ 40.89 - $ 10.22 = $ 30.67 per each lampost if you replace the incadescent bulb with the energy efficient fluorescent bulb.
Explanation:
kW.h means kilowatt.hour and is a unit of energy derived from the definition of power.
The definition of power is energy per time:
Thus, you get:
For the units, you get:
- [Energy] = [Power] [time] = KW.h.
<u>1. Calculate the energy used by the 100 W incandescent light bulb that is used 7.00 h per day, in a year (365 days).</u>
- E = Power × time = 100 W × 7.00 h / day × 365 day / year × 24 h= 255,500 W.h/year
- Divide by 1,000 to convert to kW: 255.5 kW.h/year
<u />
<u>2. Calculate the energy used by the 25 W flourescent bulb:</u>
- E = 25 W × 7.00 h / day × 365 day / year × 24 h= 63,875 W.h/year
- Divide by 1,000 to convert to kW: 63.875 kW.h/year
<u>3. Now, you can calculate both costs:</u>
- Cost per year = cost per kW.h times number of kW.h per year
<u>100 W bulb:</u>
<u />
- $ 0.16/ kW.h × 255.55 kW.h/ year = $ 40.89 / year
<u>25 W bulb:</u>
- $ 0.16/ kW.h × 63.875 kW.h/ year = $ 10.22 / year
B
If I’m correct each time that the gas increases it is going up by four meaning, each time it’s always four at the same volume