YOu should use the Graham's Law of Diffusion
R1/ R2 = √(M2/ M1)
Helium is R1, air is R2.
I'm not sure if your professor gives density or molar mass for this problem? You need M2 and M1 to solve for the rates of diffusion.
Answer:
0.095 moles of O₂ are left over.
Explanation:
First of all, state the balanced reaction:
2NO + O₂ → 2NO₂
We determine moles of each reactant:
20.2 g . 1mol / 30g = 0.673 moles of NO
13.8g . 1mol / 32g = 0.431 moles of oxygen
Oxygen is the excess reactant. Let's see.
For 2 moles of NO I need 1 mol of O₂
Then, for 0.673 moles of NO I may use (0.673 .1) /2 = 0.336 moles
I have 0.431 moles of O₂ and I only need 0.336 mol. According to reaction, stoichiometry is 2:1.
In conclussion, the moles of excess reactant that will be left over:
0.431 - 0.336 = 0.095 moles
Answer:
6.52×10⁴ GHz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 4.6 μm
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm
4.6 μm = 4.6×10¯⁶ m
Next, we shall determine frequency of the light. This can be obtained as follow:
Wavelength (λ) = 4.6×10¯⁶ m
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
v = λf
2.998×10⁸ = 4.6×10¯⁶ × f
Divide both side by 4.6×10¯⁶
f = 2.998×10⁸ / 4.6×10¯⁶
f = 6.52×10¹³ Hz
Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:
1 Hz = 1×10¯⁹ GHz
Therefore,
6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz
6.52×10¹³ Hz = 6.52×10⁴ GHz
Thus, the frequency of the light is 6.52×10⁴ GHz
Answer:
Yes
2Ag + Cu(No3)2 -----(heat) ----- Cu + 2Agno3
When heat energy is passed through Cu(No3)2 then the reverse reaction will occur
