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r-ruslan [8.4K]
2 years ago
5

Elements in group 2 are all called alkaline earth metals. What is most similar about the alkaline earth metals? how many protons

and neutrons they have which chemical properties they have how many total electrons they have which period they are most often found in
Physics
2 answers:
Rom4ik [11]2 years ago
4 0

Elements in group 2 are called alkaline earth metals, the most similarity about the alkaline metals is which chemical properties they have.

<h2>Further Explanation </h2><h3>Periodic table  </h3>
  • Periodic table is a table that contains elements arranged in columns called groups and rows called periods.
  • Elements are arranged based on physical and chemical properties such that elements in the same group will have similar physical and chemical properties.  
<h3>Chemical families  </h3>
  • Based on the chemical properties elements belong to a family of elements sharing similar chemical or physical characteristics.
  • Examples of common chemical families include; alkali metals, alkaline-earth metals, halogens and noble gases among others.
<h3>Alkaline-earth metals  </h3>
  • These are elements that are found in group 2 of the periodic table. Alkaline-earth metals include, Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.  

Properties  of Alkaline-earth metals

  • Alkaline earth metals have a valence of two since they form ions by losing two electrons from their outermost energy levels.
  • Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive compared to alkali metals.
  • Alkaline-earth metals are highly metallic and are good conductors of electricity.  
  • They react with water and steam to form metal hydroxide and metal oxides respectively  
  • They react with air to form metal oxides  
  • Reactivity of alkaline metals depends on the ease of losing electrons, thus the reactivity increases down the group as the number of energy levels increases.
  • Additionally, alkaline-earth metals have low electronegativities and low electron affinities  

Keywords: Chemical families, alkaline-earth metals, reactivity  

<h3>Learn more about:  </h3>
  • Chemical families: brainly.com/question/1358941
  • Alkaline-earth metals: brainly.com/question/8498732
  • Properties of alkaline-earth metals: brainly.com/question/11116789
  • Reactivity of metals: brainly.com/question/7101478

Level: High school  

Subject: Chemistry  

Topic: Periodic table and chemical families  

Sub-topic: Alkaline-earth metals  

Jlenok [28]2 years ago
4 0

Answer:

which chemical properties they have

Explanation:

Guest
1 year ago
edge 2021
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Answer:

<em>Entropy Change = 0.559 Times</em>

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

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2 years ago
Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
1 year ago
Read 2 more answers
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
2 years ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
1 year ago
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