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Trava [24]
4 years ago
11

Watch the animation depicting Rutherford's experiment and choose which of the following conclusions are correct.Check all that a

pply.1. The mass of an atom is concentrated at the nucleus.2. A positive charge is spread equally over the atom.3. Positive charge is condensed in one location within the atom.4. The majority of the space inside the atom is empty space.5. The atom is a very compact entity without any empty space.6. The atom contains a positively charged nucleus.
Chemistry
1 answer:
Blababa [14]4 years ago
6 0

Answer:

  • <em>1. The mass of an atom is concentrated at the nucleus.</em>

<em />

  • <em>3. Positive charge is condensed in one location within the atom.</em>

<em />

  • <em>4. The majority of the space inside the atom is empty space.</em>

<em />

  • <em>6. The atom contains a positively charged nucleus.</em>

Explanation:

When J.J Thmpson discovered the electron, he depicted the atom by the plum pudding model: a solid dough of homogeneously distributed positive charge (the pudding) containing negatively charged electrons (the plums).

Later, the scientist <em>Ernest Rutherford</em>, with its experiment of the gold foil experiment showed that the subatomic particles where not all concentrated in a solid part.

When a thin gold foil was bombarded with alpha particles (positively charged nuclei of helium atoms), most of the particles went through the gold sheet, with little deviation, but some particles bounded with a high deviation.

Such few high deviations were explained by the fact that there was a heavy region in the atom (the core or nucleus) with the positive charge that repelled the positively charged alpha particles.

Thus, <em>the mass of the atom was conentrated at the nucleus</em> (choice 1), where the positive charge is distributed in one location, which is the nucleus (not over the entire atom, just on the nucleus) discarding the choice number 2 (that a positive charge is spread equally over the atom) and proving choices 3 (<em>the positive charge is condensed in one location within the atom</em>) and 6 (<em>the atom contains a positively charged nucleus</em>).

Since most of the particles indeed went through the nucleus, this nucleus has to  occupy little space, and most of the atom was empty space, proving choice 4 (<em>the majority of the space inside the atom is empty space</em>).

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Consider the reaction
SOVA2 [1]

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

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