All categories

2 years ago

"Which of the following is true about the atom shown? Choose all that apply.

Physics

2 answers:

aalyn [17]2 years ago

8 0

Option B & Option D is your correct answers.

'cause here, atom has eight electrons in it's valence shell, so it means it has stable structure which falls in group 18

Hope this helps!

Lapatulllka [165]2 years ago

7 0

Explanation:

The given atom contains in total 18 electrons as electronic distribution is 2, 8 here. Therefore, it means that this atom belongs to group 18.

Since, the outermost shell of this atom has 8 electrons which means that it is a stable atom as it neither needs gain or lose electrons.

Hence, this atom will be unreactive in nature.

Thus, we can conclude that it is true about the atom shown that:

It is stable.

It belongs to Group 18.

You might be interested in

The last is 80 meter per second,please in am in exam

inna [77]

Answer:

I have fond the answer

Explanation:

but my camera doesn't work

7 0

3 years ago

Read 2 more answers

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

2 years ago

Anong mga katutubong pilipinong nanirahan sa kabundukan ng cordillera

Pepsi [2]

Answer:

???

Explanation:

???????????????????

7 0

2 years ago

Read 2 more answers

With which of the following is the weak nuclear force associated

BaLLatris [955]

You need to post a picture bro

5 0

2 years ago

A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial

Sveta_85 [38]

B- the student changed too many variables

7 0

2 years ago