Answer:
THE NEW VOLUME OF THE OXYGEN GAS AT 28 PSI FROM 72.5 PSI IS 0.078 L.
Explanation:
Initial volume of the oxygen in the container = 30.0 mL = 30 / 000 L = 0.03 L
Initial pressure of the oxygen = 72.5 psi = 1 psi = 6890 pascal
Final pressure = 28 psi
Final volume = unknown
First convert the mL to L and since both pressures are in similar unit that is psi; there is no need converting them to pascal or other standard unit of pressure. They cancel each other out.
This question follows Boyle's equation of gas laws and mathematically it is written as:
P1 V1 = P2 V2
Re-arranging by making P2 the subject of the formula, we have:
V2 = P1 V1 / P2
V2 = 72,5 * 0.03 / 28
V2 = 2.175 /28
V2 = 0.0776 L
The new volume of the oxygen gas at a change in pressure from 72.5 psi to 28 psi is 0.078 L.
I’m not a 100% shure but I would personally say OIL SPILLS.
Answer:
the balanced equation to the combustion of gasoline is C8H18 + 12.5 O2 → 8 CO2 + 9 H2O (1)
Answer: i would say producer
Explanation:
Answer:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
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<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
