All categories

3 years ago

Which element in the third period would you expect to have the larger atomic radius, sodium (Na) or Sulfur (S)?

Chemistry

2 answers:

jeyben [28]3 years ago

7 0

Answer: The correct answer is Option B.

Explanation:

The given elements that is sodium and sulfur are the elements which belong to the same period that is third period.

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

Therefore, the correct answer is Option B.

Doss [256]3 years ago

3 0

Answer:

Explanation:

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

You might be interested in

The melting points of alkaline earth metals are many times higher than those of the alkali metals. Explain this difference on th

7nadin3 [17]

The correct answer is higher melting point, bound by metal metal bonds.

While alkali metals only have one valence electron, alkaline earth metals have two. Metal to metal connections hold the metals together. Alkaline earth metals have a stronger metallic connection and a higher melting point because they have two valence electrons.

the characteristics that Group 2 metals excel in over Group 1 metals.

Initial Ionization Potential
Group 2 items are more difficult than group 1 elements.
Strong propensity to produce bivalent compounds

As a result, group 2 metals have stronger metallic bonding, which leads to increased cohesive energy and compact atom packing. This explains why group 2 metals are harder and have higher melting and boiling temperatures than group 1 metals.

To learn more about Group 2A(2) refer the link:

brainly.com/question/9431096

#SPJ4

3 0

2 years ago

How many electrons does phosphorous (P) need to gain to have a stable outer electron level?

Brut [27]

The correct answer is 3.

4 0

3 years ago

Read 2 more answers

How do bacteria in the guts of cows help the cows to digest the grass they eat?

Andreyy89

A cow's digestive system is quite different from that of humans. Cows eat grass, hay and other plant material that contain hard-to-digest cellulose. To cope with this they have a large stomach with four compartments, with the largest being the rumen.

3 0

3 years ago

Before a child takes a bath, his mother mixes bubbles into bath water. Physical or Chemical Change?

NeX [460]

Answer:

chemical

Explanation:

6 0

3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

3 years ago