Chemistry, 13.07.2019 08:40 KayleighMorganhopkin

The nucleus of an atom is dense and positively charged. what was observed when positively charged particles were radiated onto a gold atom during rutherford's experiment because of this? negative charges were concentrated at the center of the atom. particles that struck the center of the atom were repelled. particles that struck the edges of the atom were repelled. positive charges were distributed all over the atom.

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Answer from: jessiegarrison1

Particles that struck the center of the atom were repelled.

Explanation:

In 1911, New Zealand physicist Ernest Rutherford carried out an experiment in order to deepen his knowledge of the atomic model hitherto adopted, which was that of Thomson; in which the atom would be a sphere of positive, non-massive electrical charge, encrusted with (negative) electrons, so that its total electrical charge would be zero.

To carry out such an experiment he bombed a very thin gold leaf (thickness of approximately 10-4 mm), through a beam of alpha particles (α), coming from a sample of polonium. The polonium was inside a lead block, with an orifice, through which only the emissions of alpha particles would be allowed out.

In addition, lead plates with holes were placed in their centers, which would orient the beam towards the gold blade. And, finally, a bulkhead covered with zinc sulfide, which is a fluorescent substance, was placed behind the slide, where it was possible to visualize the path taken by the alpha particles.

At the end of this experiment, Rutherford noticed that most of the alpha particles passed through the slide, did not deflect or retreat. Some alpha particles deflected, and very few receded. In addition, the particles that reached the center of the atom were repelled.

Based on these data, Rutherford concluded that, contrary to what Dalton thought, the atom could not be massive. But in reality, much of the atom would be empty and it would contain a very small, dense and positive nucleus.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

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The nucleus of an atom is dense and positively charged. what was observed when positively charged pa...