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3 years ago

Electrons are added into the outermost what in groups 1 & 2

Physics

1 answer:

aivan3 [116]3 years ago

3 0

S orbital.

Group 1 elements have a general configuration $ns^{1}$ , where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration $2s^{1}$ whereas Cesium has $6s^{1}$ . Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of $ns^{2}$ . For example, Magnesium has $3s^{2}$ as its outer shell configuration while Strontium has the same as $5s^{2}$ .

We see that in both the cases, the outermost S orbital is being filled.

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ (we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why.

user100 [1]

Answer:

The ballon would be inflated. The reason is that the sodium bicarbonate in baking soda reacts with acetic acid in vinegar to produce gas.

Explanation:

The main component of baking soda is sodium bicarbonate, ${\rm Na_{2}CO_{3}}$ .

Vinegar is mostly a solution of acetic acid ${\rm CH_{3}COOH}$ in water.

Acids such as acetic acid react with carbonate salts. One of the products of such reactions is carbon dioxide ${\rm CO_{2}}$ , a gas.

In this question, when the acetic acid in vinegar reacts with sodium bicarbonate in the baking soda, the following reaction would occur:

$\begin{aligned}& {\rm Na_{2}CO_{3}}\, (aq) + 2\, {\rm CH_{3}COOH}\, (aq) \\ &\to 2\, {\rm CH_{3}COONa}\, (aq) + {\rm CO_{2}}\, (g)\end{aligned}$ .

The ${\rm CO_{2}}$ produced would then inflate the ballon placed on the opening of the bottle.

5 0

2 years ago

The pressure in Arabian sea is much bigger than near the surface? please don't scam

sashaice [31]

Answer:

In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing

stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.

Explanation:

4 0

3 years ago

Read 2 more answers

LIst properties common to many organic compounds

stich3 [128]

Some properties I know are melting point, boiling point, and index of refraction.

7 0

3 years ago

What is used to measure mass?

defon

Answer:

A balance

Explanation:

Scientists measure mass with a balance, such as a triple beam balance or electronic balance. In science, the volume of a liquid might be measured with a graduated cylinder.

4 0

3 years ago

Read 2 more answers