1. Radiation with a frequency of 1.85 MHz, falls in the radio wave region. What is the wavelength (in nm) of this electromagnetic radiation? (HINT: Hz is "cycles per second" and should be replaced by 1/s as a unit in the calculation. First convert MHz to Hz (106) also the final answer convert to nm. (10-9)

2. What is the frequency of visible light, which has a wavelength of 625 nm? What color is the wave?

(HINT: Notice for final answer that nm must be converted to m in order for the units to cancel properly)

1. 1.62×10¹¹ nm

2. 4.8×10¹⁴ Hz

Explanation:

1. Determination of the wavelength.

We'll begin by converting 1.85 MHz to Hz. This can be obtained as follow:

1 MHz = 10⁶ Hz

Therefore,

1.85 MHz = 1.85 MHz × 10⁶ Hz / 1 MHz

1.85 MHz = 1.85×10⁶ Hz

Next, we shall determine the wavelength of the electromagnetic radiation. This can be obtained as follow:

Frequency (f) = 1.85×10⁶ Hz

Velocity of electromagnetic radiation (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 1.85×10⁶

Divide both side by 1.85×10⁶

λ = 3×10⁸ / 1.85×10⁶

λ = 162.16 m

Finally, we shall convert 162.16 m to nm. This can be obtained as follow:

1 m = 10⁹ nm

Therefore,

162.16 m = 162.16 m × 10⁹ nm / 1 m

162.16 m = 1.62×10¹¹ nm

Thus, the wavelength of the electromagnetic radiation is 1.62×10¹¹ nm

2. Determination of the frequency.

We'll begin by converting 625 nm to m. This can be obtained as follow:

1 nm = 10¯⁹ m

Therefore,

625 nm = 625 nm × 10¯⁹ m / 1 nm

625 nm = 6.25×10¯⁷ m

Finally, we shall determine the frequency the visible light as follow:

Wavelength (λ) = 6.25×10¯⁷ m

Velocity of light (v) = 3×10⁸ m/s

Frequency (f) =?

v = λf

3×10⁸ = 6.25×10¯⁷ × f

Divide both side by 6.25×10¯⁷

f = 3×10⁸ / 6.25× 10¯⁷

f = 4.8×10¹⁴ Hz

Therefore, the frequency of the visible light is 4.8×10¹⁴ Hz

the answer is barium and copper bromide

the formula mass of a molecule or a compound is the sum of the atomic weights of the atoms in its empirical formula

(that's what i know about formula mass)