All categories

3 years ago

Compare and contrast properties of halogens in group 17 with the alkali metal family in group 1

Physics

1 answer:

N76 [4]3 years ago

7 0

Answer:

See the explanation below.

Explanation:

Halogens are compound of Group 17 of the periodic table. It's properties are:

Down the group the melting and boiling point of the elements increases, so, Fluorine will have the lowest whereas, Astatine will have the highest.
Down the group reactivity of halogens decreases, Fluorine reacts instantly whereas, iodine need to be heated strongly for reaction to take place.

Alkali metals are compound of Group 1 of the periodic table. It's properties include:

Alkali metal have low melting point and down the group melting point decreases i.e. Lithium have the melting point of 180 degree Celsius whereas, Cesium have the melting point of 28 degree Celsius only.
Down the group reactivity of these metal increases For example, Cesium reacts with Chlorine more vigorously as compare to reaction of Potassium with Chlorine.

You might be interested in

Scientific theories have strod the test of time and will not change

Nataly [62]

That is not a question but not all scientific theories have stood the test of time

8 0

3 years ago

What is matter explain verifly

marysya [2.9K]

Answer:

Matter is a substance that has inertia and occupies physical space. According to modern physics, matter consists of various types of particles, each with mass and size.Matter can exist in several states, also called phases. The three most common states are known as solid, liquid and gas.Matter is the Stuff Around You or Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).Solid ice, water and steam are few examples of matter touched in everyday life. Subatomic particles are also considered as matter.

6 0

3 years ago

Read 2 more answers

how could a person easily tell which letters had four and which rive pages without opening the envelope? what assumptions did yo

Lubov Fominskaja [6]

Had 4 what and when you said rive did u mean 5

5 0

3 years ago

The axle of an automobile is acted upon by the forces and couple shown. knowing that the diameter of the solid axle is 32 mm, de

saul85 [17]

First we need to convert the mm to inches to make our computation easier.

1mm = 0.0393701

32mm * 0.0393701 = 1.25 in

Solution:

C = 1/2d = ½ (1.25) = 0.625 in^4

Tension: tension = Te/J = 2T/ piC^3

= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi

Bending:

I = pi/4 * c^4 = 119.842 x 10^-3 in^4

M = (5)(600) = 3600 lb in

G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2 psi = -18.775ksi

Gx = -18.775 ksi

Gy = 0

Txy = 6.519 ksi

G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

1. G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi

G2 = Gave - R = -9/387 - 11.429 = -20.8

Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 = -1.3889

ϴp = -27.1 degrees and 62.9 degrees

2. Tmax = R = 11.43 ksi

R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

3 0

3 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago