Hydrogen atoms do not repel each other within a star

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig

m_a_m_a [10]

Answer:

Explanation:

Let's define positive to be right. The net force on q2 is the

vector sum of the forces from q1 and q3.

The force has a magnitude of and points to the left.

Handling the signs correctly is critical. The negative signs

come from the direction of each of the forces (both to the left),

not from the signs of the charge.

2. Both charges 1 and 3 have 1 close

neighbor and 1 far neighbor. Will

they experience forces of the same

magnitude?

No, because both forces acting on charge 1 are in the same

direction, while the two forces acting on charge 3 are in

opposite directions. Thus, 1 > 3.

Both charges 1 and 2 have two forces acting in the same

direction. Will they experience forces of the same magnitude?

No, because one force acting on charge 1 is the same

magnitude as one acing on charge 2, while the second force

acting on charge 1 is smaller – it comes from a charge farther away

3 0

4 years ago

How did Niels Bohr describe electrons in his atomic model? Their energies can have any values. Their exact positions cannot be k

Fed [463]

Answer is: They orbit the central nucleus in discrete paths.

According to the Bohr model of the atom:

1. Electrons orbit the nucleus in orbits that have a set size and energy.

2. Energy levels of electrons are discrete (certain discrete values of energy).

3. Electrons can jump from one energy level to another, absorbing or emitting electromagnetic radiation with a frequency ν (energy difference of the levels).

5 0

3 years ago

Read 2 more answers

Why does ice float after it crystallizes?

Vesnalui [34]

C. Its density is less than water
This is why. The denser the object is the farther it sinks. Since crystallized ice floats it is less dense than the surrounding liquid.

4 0

4 years ago

Read 2 more answers

Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a

Alexandra [31]

Answer:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

Explanation:

1. Balanced chemical equation (given):

$2A(aq)+2B(aq)\rightleftharpoons 5C(aq)$

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

$K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

a. Increase [B]

Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant $K_{c}$ is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

b. Increase [A]

Rightward shift.

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

Leftward shift.

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for $K_{c}$ : by increasing the denominator the numerator has to increase to keep the same value of $K_{c}$

d. Decrease [A]

Leftward shift.

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

Leftward shift.

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

Rightward shift.

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

No shift

You need to perform some calculations and determine the reaction coefficient, $Q_c$ to compare with the equilibrium constant $K_{c}$ .

The expression for $Q_c$ has the same form of the equation for $K_{c}$ . but the it uses the inital concentrations instead of the equilibrium concentrations.

$Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

Doubling [A] and reducing [B] to one half would leave the product of [A]² by [B]² unchanged, thus $Q_c$ will be equal to $K_{c}$ .

When $Q_c$ = $K_{c}$ the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

Rightward shift.

Again, using the expression for $Q_c$ , you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that $Q_c$ will be greater than $K_{c}$ ..

When $Q_c$ > $K_{c}$ the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

Summarizing:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

4 0

4 years ago

Which one is true and which one is false

Mekhanik [1.2K]

In order top top to bottom
F
F
T
T
F
F
T
T

7 0

4 years ago