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3 years ago

A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?

Physics

1 answer:

ASHA 777 [7]3 years ago

5 0

The complete question is: A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?

A) The drawing is wrong because there are more chemicals on the products side.

B) The drawing is correct because there are 12 compounds on each side of the arrow.

C) The drawing is wrong because there are different compounds on each side of the arrow.

D) The drawing is correct because there are 12 atoms of each type on each side of the arrow.

Answer:

Option D is correct

Explanation:

In the diagram attached below, it can be seen that there are 12 atoms of element which combine with 12 atoms of another element forming a compound. For the drawing to be correct, there should be 12 atoms of each type of element on both the reactants as well as product side, which is the case. There cannot be imbalance in the number of atoms of different elements on the two sides for a chemical reaction to occur.

Hence, option D is correct.

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The force generated by a single muscle fiber can be increased by increasing is called

mars1129 [50]

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

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3 years ago

If the soccer ball has a mass of 284g, and gravitational force is 9.81 m/s2 on Earth, what is the weight of the soccer ball?

bagirrra123 [75]

Answer:

2.79 N
8.34 N
Watermelon

Explanation:

Weight of soccer ball

Weight = mass (in kg) × gravitational force
Weight = 284 x 10⁻³ x 9.81
Weight = 0.284 x 9.81
Weight = 2.79 N (approximately)

Weight of watermelon

Weight = mass (in kg) x gravitational force
Weight = 850 x 10⁻³ x 9.81
Weight = 0.85 x 9.81
Weight = 8.34 N (approximately)

As the watermelon has more weight, it will hit the ground first.

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2 years ago

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic f

alekssr [168]

Answer:c

Explanation:

When the direction of current is towards the observer then the magnetic field around it will be in the form of concentric circles and its direction will be anti-clockwise when viewed from the observer side.

Whenever current is flowing in a current-carrying conductor then the magnetic field is associated with it and direction of the magnetic field is given by right-hand thumb rule according to which if thumb represents the direction of current then wrapping of fingers will give the direction of the magnetic field

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3 years ago

A mass m1 hangs from a spring k and is in static equilibrium. A second mass m2 drops through a height h and sticks to m1 without

jekas [21]

Answer:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$

Explanation:

From the information given:

The equation of the motion can be represented as:

$(m_1 +m_2) \hat u + ku = m_2 g--- (1)$

where:

$m_1$ = mass of the body 1

$m_2$ = mass of the body 2

$\hat u$ = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency $\omega _n = \sqrt{\dfrac{k}{m_1+m_2}}$

And the equation for the general solution can be represented as:

$u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)$

To determine the initial velocity, we have:

$\hat u_2^2 = 2gh$

$\hat u_2 = \sqrt{2gh}$

where h = height

Suppose we differentiate equation (2) with respect to time t; we have the following illustration:

$\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0$

now if t = 0

Then

$\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0$

$= \omega _n B$

Using the law of conservation of momentum on the impact;

$m_2 \hat u_2=(m_1+m_2) \hat u (0)$

By replacing the value of $\hat u_2$ with $\sqrt{2gh$

Then the above equation becomes:

$m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)$

Making u(0) the subject of the formula, we have:

$u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}$

Similarly, the value of the variable can be determined as follows;

Using boundary conditions

$0 = A cos 0 + B sin 0 + \dfrac{m_2g}{k}$

$0 = A (1)+0+ \dfrac{m_2g}{k}$

$A =- \dfrac{m_2g}{k}$

Also, if $\hat u (0) = \omega_nB$

Then :

$\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B$

making B the subject; we have:

$B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}$

Finally, replacing the value of A and B back to the general solution at equation (2); we have the equation of the subsequent motion u(t) which is:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$

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3 years ago

Under what conditions does the magnitude of the average velocity equal the average speed?

Dmitry [639]

So when the body moves in a straight line the average velocity is equal to the average speed. average velocity would be equal to average speed when the total distance travelled equals the net displacement of a particle. this happens when a particle moves along a straight line in a fixed direction.

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3 years ago