Answer:
Explanation:
Since temperature remains constant, the only variables that change are volume and pressure. Therefore, we are using <u>Boyle's Law.</u> This states that the pressure is inversely proportional to the volume. The formula is:
We know the gas starts with a volume of 20.0 liters at a pressure of 275 kPa. We can substitute these values into the left side of the formula.
We know the gas expands to a volume of 35.0 Liters, but we do not know the pressure.
Since we are solving for the new pressure, we must isolate the variable P₂. It is being multiplied by 35.0 Liters and the inverse of multiplication is division. Divide both sides by 35.0 L.
The units of liters cancel.
The original measurements of pressure and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.
The 1 in the tenths place (157.142857) tells us to leave the 7 in the ones place.
If the gas expanded to a volume of 35.0 liters while the temperature remained constant, the pressure in container was approximately 157 kilopascals.
Here is your answer:
The amount of electrons that the element Beryllium has is: 4
Reason:
Beryllium atomic number is 4
Therefore the amount of protons and electrons Beryllium has is 4
Your answer is 5.
Hope this helps.
According to the molar gas volume, one mole of a gas occupies a volume of 22400 cm³ or 22.4 liters. The standard temperatures and pressures conditions are temperature of 0 degrees Celsius or 273 kelvin and an absolute pressure of 103.325 kPa or 1 atm. In this case, the density is 0.902 g/l and the volume is 22400 cm³,
therefore, the molar mass will be;
0.902 g/ L ×22.4 liters = 20.2048g
Answer:
a) The work done is 10.0777 kJ
b) The water's change in internal energy is -122.1973 kJ
Explanation:
Given data:
1 mol of liquid water
T₁ = temperature = 100.9°C
P = pressure = 1 atm
Endothermic reaction
T₂ = temperature = 100°C
1 mol of water vapor
VL = volume of liquid water = 18.8 mL = 0.0188 L
VG = volume of water vapor = 30.62 L
3.25 moles of liquid water vaporizes
Q = heat added to the system = -40.7 kJ
Questions: a) Calculate the work done on or by the system, W = ?
b) Calculate the water's change in internal energy, ΔU = ?
Heat for 3.25 moles:
The work done:
The change in internal energy:
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