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4 years ago

How do you balance this equation Mg+ O2= MgO

Physics

1 answer:

Anuta_ua [19.1K]4 years ago

8 0

Answer:

2 Mg(s) + O2(g) → 2 MgO(s)

As magnesium (Mg) has a valency of +2, and oxygen (O) has a valency of -2, the ratio would be 1:1, and magnesium oxide would be represented as MgO. It is insoluble in water, hence it has the subscript as a solid, represented by (s). In order to use up the diatomic oxygen (O2), there needs to be two moles of magnesium (2Mg) on the reactant side. This would produce 2 moles of MgO on the product side.

Explanation:

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Laser light has a great deal of______ a. chemical b. heat c. potential d. sound energy.??

svet-max [94.6K]

Answer:

b. heat

Explanation:

4 0

3 years ago

How much kinetic energy does a 6kg cart have when moving at 4 m/s?

Arturiano [62]

Given:

mass = 6 kg

velocity = 4 m/s

To find:

Kinetic energy of the cart = ?

Formula used:

Kinetic energy = $\frac{1}{2} m v^{2}$

Where m = mass of the cart

v = velocity with which the cart is moving

Solution:

Kinetic energy of the moving cart is given by,

Kinetic energy = $\frac{1}{2} m v^{2}$

Where m = mass of the cart

v = velocity with which the cart is moving

Kinetic energy = $\frac{1}{2} \times 6 \times \ 4 \times 4$

Kinetic energy = 48 Joule

Thus, the kinetic energy of the moving cart is 48 Joule.

3 0

4 years ago

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!

BARSIC [14]

<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.

Now, in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at $9.8\frac{m}{s^{2}}$ and hit the ground at approximately the same time.

5 0

4 years ago

A loop circuit has a resistance of R1 and a current of 1.8 A. The current is reduced to 1.6 A when an additional 3.8 Ω resistor

Allisa [31]

Answer:

Value of $R_1=30.4ohm$

Explanation:

We have given

In first case resistance is $R_1$ and current is 1.8 A

Let the potential difference is v

So $1.8=\frac{v}{R_1}$ ----eqn 1

In second case resistance is $R_1+3.8$ and current is 1.6 A and potential difference will be as it is a series connection

So $1.6=\frac{v}{R+3.8}$ ----eqn 2

From eqn 1 and eqn 2

$1.8R_1=1.6R_1+6.08$

$R_1=30.4ohm$

6 0

3 years ago

Read 2 more answers

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago