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:
i think the answer is D. energy
Explanation:
To determine which is the correct answer, we convert the mass of the compounds into units of particles of the compound. We use the molar masses and the Avogadro's number. We do as follows:
<span>76.9 g I2 (1 mol / 253.81 g) (6.022x10^23 particles / 1 mol ) = 1.82x10^23
79.9 g Br2 (1 mol / 159.81 g)</span>(6.022x10^23 particles / 1 mol ) = 3.011x10^23<span>
6 g C (1 mol / 12 g)</span>(6.022x10^23 particles / 1 mol ) = 3.011x10^23<span>
13.01 g CH4 ( 1 mol / 16.04 g )</span>(6.022x10^23 particles / 1 mol ) = 4.88x10^23 particles
Therefore, the answers are Br2 and C.
