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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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Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?

Neko [114]

Answer:

<h3>The answer is 0.54 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{10.5}{19.3} \\ = 0.54404145...$

We have the final answer as

Hope this helps you

7 0

2 years ago

How is energy measured?

stepladder [879]

Because a Btu is so small, energy is usually measured in millions of Btus. 1 Btu = the amount of energy required to increase the temperature of one pound of water (which is equivalent to one pint) by one degree Fahrenheit. This is roughly the heat produced from burning one match.

<em>https://www.ucsusa.org/clean_energy/our-energy-choices/how-is-energy-measured.html</em>

5 0

2 years ago

For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?

Reptile [31]

Answer: Gravitational force and drag force

Explanation:

For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,

4 0

2 years ago

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

Alekssandra [29.7K]

Answer:

$0.53m/s^2$

Explanation:

We are given that

Mass of wire=m=3.6 g= $3.6\times 10^{-3} kg$

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms= $60.1\times 10^{-3} s$

$1 ms=10^{-3} s$

Speed,v= $\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s$

$g=\frac{v^2m}{m'l}$

Using the formula

$g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2$

8 0

3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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6 0

1 year ago