Answer:
B
Explanation:
Bad conductor of elecricity
Answer:
The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J
Explanation:
The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).
Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

So:

In this case, the wavelength is 3.35mm=3.35*10⁻³m and the energy per photon is:

E=5.93*10⁻²³ 
The detector is capturing 3.3*10⁸ photons per second. So, in 1 hour:

E=7.04*10⁻¹¹ 
The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J
Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)