Chemistry, 12.08.2020 04:01 yialefi

Os subníveis mais energéticos de um dado átomo são: ...4s2 3d10 4p5 a) indique o seu número atomico b) quantos electrões de valência apresenta esse átomo c) a que família pertence?

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Answer from: mstrish71oteauw

A. 35

B. 7

C. halogênios

Explanation:

Aqui, para responder a essa pergunta, precisaremos conhecer o elemento particular em questão.

4s ^ 2 3d ^ 10 4p ^ 5 significa que está a cinco elétrons da configuração eletrônica do último elemento na primeira camada dos metais pesados.

O último elemento da 1ª série do elemento de transição é o zinco, portanto, como está a apenas 5 elementos de distância, o átomo de que estamos falando é o átomo de Bromo de Bromo.

A. O zinco tem um número atômico 30 e como o bromo está a 5 elementos de distância, seu número atômico é 35

B. Uma vez que pertence ao grupo halogênio, tem 7 elétrons de valência como o resto da família

C. Pertence à família dos halogênios

Answer from: Quest

answer: the figure is attached herewith.

step-by-step explanation: we are given to sketch a unit circle and label the terminal points corresponding to

$\theta=\dfrac{\pi}{3},~2\dfrac{\pi}{3},~3\dfrac{\pi}{3},~4\dfrac{\pi}{3},~5\dfrac{\pi}{3}.\\\\\\\rightarrow \theta=60^\circ,12^\circ,180^\circ,240^\circ,300^\circ.$

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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Os subníveis mais energéticos de um dado átomo são: ...4s2 3d10 4p5 a) indique o seu número atomico...

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