Explanation:
the volume and temperature of a gas have a ditect relationship,as the temperature increases the volume also increases when pressure is held constand, heating the gas increases the kinetic energy of the particles or atoms,causing the gas to expand until the pressure returns to its original value
Answer:
1/3p0
Explanation:
The combined gas law:
P1V1/T1 = P2V2/T2, where P, V and T are Pressure, Volume, and Temperature. Temperature must always be in Kelvin. The subscriopts 1 and 2 are for initial (1) and final (2) conditions.
In this case, temperature is constant (adiabatically). V1 = 2.0L and V2 = 6.0L. I'll assume P1 = p0.
Rearrange the combined gas law to solve for final pressure, P2:
P1V1/T1 = P2V2/T2
P2 = P1*(V1/V2)*(T2/T1) [Note how I've arranged the volume and temoperature terms - as ratios. This helps us understand what the impact of raising or lowering one on the variables will do to the system].
No enter the data:
P2 = P1*(V1/V2)*(T2/T1): [Since T2 = T1, the (T2/T1) terms cancels to 1.]
P2 = p0*(2.0L/6.0L)*(1)
P2 = (1/3)p0
The final pressure is 1/3 the initial pressure.
It can be represented by atom -9
It would be Fluorine in real..
Water has greatest boiling and melting points because strong hydrogen bonds between water molecules (intermolecular force). That is because oxygen has largest electronegativity and smallest atomic radius of all elements in group. Other elements in this group do not form hydrogen bonds and they are gasses in room temperature, water is liquid and their boiling and melting points are lower. Boiling and melting point of hydrides gradually increases from H₂S to H₂Te because of increase of the radius and molecular mass of atoms from sulfur to tellurium, that is why intermolecular force of attraction of molecules increases. <span>
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The question is incomplete and complete question is :
It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.050 mL. Assuming the density of liquid water is 1.00 g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100°C and 1atm ?
Answer:
0.0028 moles of water are in one drop of liquid.
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.
Explanation:
Volume of of drop = v = 0.050 mL
Mass of drop = m
Density of water = d = 1.00 g/mL

Moles of water in drops:

0.0028 moles of water are in one drop of liquid.
Pressure at which 0.0028 moles are vaporized = P = 1 atm
Temperature at which 0.0028 moles are vaporized = T = 100°C = 100+273 K = 373 K atm
Volume of moles of water = V
Moles of water = n = 0.0028 mol
( ideal gas equation )

V = 0.086 L = 86 mL ( 1L = 1000 mL)
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.