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BTW reduction is the action or fact of making a specified thing smaller or less in amount, degree, or size.
If the temperature of a liquid-vapor system at equilibrium increases, it will shift towards the vapor phase, assuming that the pressure remains equal. The concentration of vapor will also increase relative to the concentration of liquid in the system. Thus, the new equilibrium condition will have more vapor than liquid.
Answer : Option D) The particles move enough that they are not fixed in place, and the liquid can flow.
Explanation : The kinetic energy of the particles are allowed to move freely and are in motion when in the liquid state whereas the intermolecular particles can just flow; as the intermolecular attractions between the particles allows the liquid to flow by giving them a force to flow.
Answer:
237.8L of water would need to be added.
Explanation:
The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).
Then figure out what information you have and what is being found. In this case:
M1 = 54.7 M
V1 = 1092 mL = 1.092 L
M2 = 0.25 M
V2 = unknown
Then solve the equation for whatever you are trying to find.
M1V1=M2V2
V2=M1V1/M2
Now you need to plug everything in.
V2=(54.7M*1.091L)/0.25M
V2=238.93L
That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.
238.93L - 1.091L = 237.8L
Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.
I hope this helps. Let me know if anything is unclear.
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
