Which reaction is an example of an oxidation-reduction reaction?

Tom [10]

Answer:

Radiation effects on electrical equipment depend on the equipment and on the type of ionizing radiation to which it is exposed.

First, beta radiation has little, if any, effect on electrical equipment because this type of ionizing radiation is easily shielded. The equipment housing and the construction of the parts within the housing will protect the equipment from beta-radiation (high-energy electrons) exposure.

Gamma radiation is penetrating and can affect most electrical equipment. Simple equipment (like motors, switches, incandescent lights, wiring, and solenoids) is very radiation resistant and may never show any radiation effects, even after a very large radiation exposure. Diodes and computer chips (electronics) are much more sensitive to gamma radiation. To give you a comparison of effects, it takes a radiation dose of about 5 Sv to cause death to most people. Diodes and computer chips will show very little functional detriment up to about 50 to 100 Sv. Also, some electronics can be "hardened" (made to be not affected as much by larger gamma radiation doses) by providing shielding or by selecting radiation-resistant materials.

Some electronics do exhibit a recovery after being exposed to gamma radiation, after the radiation is stopped. But the recovery is hardly ever back to 100% functionality. Also, if the electronics are exposed to gamma radiation while unpowered, the gamma radiation effects are less.

Ionizing radiation breaks down the materials within the electrical equipment. For example, when wiring is exposed to gamma rays, no change is noticed until the wiring is flexed or bent. The wire's insulation becomes brittle and will break and may cause shorts in the equipment. The effect on diodes and computer chips is a bit more complex. The gamma rays disrupt the crystalline nature of the inside of the electronic component. Its function is degraded and then fails as more gamma radiation exposure is received by the electronic component.

Gamma rays do not affect the signals within the device or the signals received by the device. Nonionizing radiation (like radio signals, microwaves, and electromagnetic pulses) DO mess with the signals within and received by the device. I put a cheap electronic game in my microwave oven at home. It arced and sparked and was totally ruined. I didn’t waste any more of my time playing that game.

Hope this helps.

Explanation:

MARK ME AS BARINIEST PLS

5 0

4 years ago

Which type of circuit has a string of lights and one light goes out but the others stay lit

kondor19780726 [428]

Answer:parrallel

Explanation:

8 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)

Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0

3 years ago

5. What happens to the average kinetic energy of water molecules as water freezeslt decreases as the water absorbs energy from i

4vir4ik [10]

The correct answer among the choices listed above is option D. The average kinetic energy of water molecules as water freeze decreases as water releases energy to its surroundings. Energy is released as the molecules go into a more condensed phase which is the solid.

3 0

3 years ago

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