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3 years ago

What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?

Physics

2 answers:

patriot [66]3 years ago

8 0

Answer:

the third option

Explanation:

igomit [66]3 years ago

4 0

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

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Two immersion heaters, A and B, are both connected to a 120.0-V supply. Heater A can raise the temperature of 1.00 L of water fr

Inessa05 [86]

Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80

Explanation:

Consider C be the specific heat of water, R₁ and R₂ be the resistance of heater A and heater B respectively.

Given:

Mass of water in heater A, m₁ = 1 L

Mass of water in heater B, m₂ = 5.80 L

Initial temperature, T₀ = 20 ⁰C

Final temperature, T₁ = 90 ⁰C

Time, t = 5 min

Amount of heat required to raise the temperature of water by heaters A and B are given by:

Q₁ = m₁C(T₁ - T₀) and

Q₂ = m₂C(T₁ - T₀)

Ratio of power used by both the heaters A and B is:

$\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }$

Since, time t, temperature difference(T₁ - T₀) and specific heat C are same for both the heaters A and B. So, the above equation becomes:

$\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }$ ...(1)

The relation to determine electrical power for both heaters A and B are:

$P_{1}=\frac{V^{2} }{R_{1} }$ and

$P_{2}=\frac{V^{2} }{R_{2} }$

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

$\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }$ ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

$\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }$

Substitute the suitable values in the above equation.

$\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }$

$\frac{R_{1} }{R_{2} }=5.80$

6 0

3 years ago

Pretend a person can walk only at a certain pace – no faster, no slower. If you time her uninterrupted walk across a room of kno

Free_Kalibri [48]

Explanation:

If a person is set to walk on a constant speed regardless of the situations then if the person walks a certain distance with no interruptions in an observed time then her speed can be calculated.

When the same person walking through the room momentarily stops to introduce herself then the average speed of the of the person slows down as it happens with the light wave when passes through glass which is an optically denser medium than the air, but the light wave does not stop anywhere in the medium.

It can be more relevant to the person's speed when she walks wading through the water or the person walks through the sand then the person feels resistance in the sand or in water which reduces her overall speed.

4 0

3 years ago

What are advantage of frictional force?

kogti [31]

The force of friction acts in the opposite direction that the object is moving, so friction slows down the motion of moving objects. Friction also produces unnecessary heat. For example, car engines become hotter because of friction.

3 0

3 years ago

Florida is a peninsula bound by the Atlantic Ocean on the east and the Gulf of Mexico on the west. The sea affects temperatures

Andreyy89

They are usually warmer

6 0

3 years ago

For the following types radio waves infared radiation visible light ultraviolet radiation gamma radiation of electromagnetic rad

ddd [48]

Answer: Option 5.

Decreases increases increases.

Explanation:

Electromagnetic spectrum refers to differences in frequencies, wavelengths and photon energies of electromagnetic radiation.

From the top of the list to the bottom i.e from radio waves, infrared waves, visible lights, ultraviolet radiation, gamma radiation, wavelengths decreases, the frequency increases as the photon energy increases.

The electromagnetic radiations : radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays have high-frequency and short wavelengths. Their wavelengths decrease. The photon energy is directly proportional to wave frequency.

6 0

4 years ago