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3 years ago

The laws of electrostatics predict that protons will

1 answer:

3 0

The laws of the electrostatics were developed by Coulomb and are further detailed or subdivided into two as follows:

(1) The first one talks about the similarities in the charges of the particles and what happens to them. This states that like charges repel.

(2) The second one quantifies the field between two particles with equation showing that field is directly proportional to the masses and inversely to the square of the distance between them.

In this item, we can then say that the protons (having the same charges) will have to repel from each other in accordance to statement 1 of the law.

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Students are completing a table about a particular subatomic particle that helps make up an atom. The students have filled in on

Semmy [17]

Answer:
<span> Its location is in the nucleus, because the particle is a proton or a neutron.</span>

5 0

3 years ago

Read 2 more answers

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Bond [772]

Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2) $^{t/t(1/2)}$ where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days

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3 years ago

4. Why can't the subscripts be changed in a chemical equation in chemistry

neonofarm [45]

If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is

4 0

2 years ago

What factors affect the dynamic state of equilibrium in a chemical reaction and how?

yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant $K_c$ . The system will shift in response to certain external shocks. At the new equilibrium $Q$ will still be equal to $K_c$ , but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of $Q$ . For most reversible reactions:

External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

Changes in pressure due to volume changes.

Catalysts do not influence the value of $Q$ . See explanation.

Explanation:

$\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}$ .

Similar to the rate constant, the equilibrium constant $K_c$ depends only on:

$\Delta G$ the standard Gibbs energy change of the reaction, and
$T$ the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium $Q = K_c$ the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

Changes in concentration influence the number of particles per unit space.
Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0

3 years ago

For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and th

Scilla [17]

Answer:

Q = 0.50

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

$Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}$

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = $\frac{[NO_2]^2}{[N_2O_4]}$

$Q = \frac{0.50^2}{0.50}$

Q = 0.50

So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.

6 0

3 years ago