Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
<em></em>
Temperature is related to the kinetic energy the atoms within any given
substance. Within a solid, we can therefore assume that the density will
increase as the solid is cooled to a low temperature, as this will
cause a loss in kinetic energy and the atoms won't be able to move as
freely.
Answer:
28.28 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have two different values of V and P:
<em>P₁V₁ = P₂V₂</em>
<em></em>
P₁ = 700.0 mm Hg, V₁ = 4.0 L.
at burst: P₂ = 99.0 mm Hg, V₂ = ??? L.
<em>∴ V₂ = P₁V₁/P₂</em> = (700.0 mm Hg)(4.0 L)/(99.0 mm Hg) = <em>28.28 L.</em>
The key to most "how do I separate." questions is solubility.
The trick is to add a liquid that will only dissolve one substance but not another.
Let's say you had a beaker full of sand, table salt (NaCl), and acetanilide. Is there anything you can add that would only dissolve one of these three substances?
Yes, there is! Acetanilide like most organic compounds, isn't soluble in water. But salt is soluble in water. So to the mixture, I would add water, and then pass the water through a filter. The filter paper will "catch" the sand and acetanilide, but the table salt will remain dissolved in the water. If you then let that water evaporate (either via boiling or under vacuum), you will recover your salt.
So now, how to do you separate the sand from the acetanilide? Sand isn't really soluble in anything, but acetanilide is soluble in organic solvents, such as ethanol. So to the mixture of sand and acetanilide, add ethanol, and pass it through a filter. The sand will once again get stuck in the filter paper, and your acetanilide will be dissolved in ethanol. Remove the ethanol (via vacuum, or rotovap) and you will be left with acetanilide.
λ=Wavelength = .129 m = 0.129 m
E = energy = ?
h =Planck constant = 6.63 × 10^-34 J.s
f = frequency of photo/ electromagnetic radiation = c/λ
c = Speed of light in a vacuum = 3 × 10^8 m/s
___o___o___
I hope I helped you^_^
