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Alex777 [14]
3 years ago
7

How much faster does helium escape through a porous container than ozone?

Chemistry
1 answer:
otez555 [7]3 years ago
4 0
According to Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".

                                r₁ / r₂  =  \sqrt{M2 / M1}   ---- (1)

r₁    =  Rate of effusion of He

r₂    =  Rate of Effusion of O₃

M₁  =  Molecular Mass of He  =  4 g/mol

M₂  =  Molecular Mass of O₃  =  48 g/mol

Putting values in eq. 1,

                                r₁ / r₂  =  \sqrt{48 / 4}

                                r₁ / r₂  =  \sqrt{12}

                                r₁ / r₂  =  3.46

Result:
          Therefore, Helium will effuse 3.46 times more faster than Ozone.
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Dr. Terror has developed a new alloy called Ultranomium. He is testing a bar that is 1.25 m1.25 m long and has a mass of 382 g.3
MrRissso [65]

Answer:0.8742j/g°C

Explanation: SOLUTION

GIVEN

length of bar=1.25m

mass 382g

temperature= 20°C to 288°C

Q=89300J

Specific Heat Capacity  will be calculated using

Q=mC∆T

where

C = specific heat capacity

Q = heat

m = mass

Δ T = change in temperature

C=Q/ m∆T

=89300/382X(288-20.6)

=0.8742j/g°C

4 0
3 years ago
if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0
3 years ago
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jok3333 [9.3K]
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2 years ago
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Explanation:

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6 0
3 years ago
Read 2 more answers
What is the equilibrium constant of pure water at 25°C?
ANEK [815]
The equilibrium reaction, causes the water dissociation constant, Kw, is 1.01 × 10-14<span> at 25 °C. That is because every H</span>+<span> (H</span>3O+) ion these forms accompanied by the formation of an OH-<span> ion, are the concentrations of these ions and in pure water the same thing can be calculated from </span>Kw<span>. 

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6 0
3 years ago
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