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4 years ago

Give an example of ionic and covalent bond, and explain and make an example for each.

Physics

1 answer:

Shtirlitz [24]4 years ago

7 0

Well, ionic bonding is the type of bonding in which electrons are transferred from one atom to another. While covalent bonding is the type of bonding in which the electrons are shared between the two atoms of the elements. An example of ionic bonding is sodium and fluorine. Sodium transfers one electron to flourine which has 7 outer shell electrons. The electrons are then made 8, which is complete. Sodium has one outer shell electron which cannot stay alone, so is transferred.
An example of covalent bonding is between two hydrogen atoms to form H₂
The two atoms come together and combine, they are attached to each other.

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A person observes that in the winter, his hair stands on end when it is brushed. This is because his hair becomes positively cha

Shalnov [3]

Answer:

Because of the transfer of electrons from hairs to the brush.

Explanation:

In winters, the hairs on brushing usually stands on the ends because when the hairs are rubbed by the brush separation of electrons take place from the parent atoms' nuclei on the hairs and thus the hairs are left positively charged.

Normally the charges are stay together and rectify the effect of each other as one is positive and the other is negative thus making the atoms neutral.

The process due to which the hairs acquire positive charge is known as charging by friction.

3 0

3 years ago

The pressure of the earth's atmosphere at sea level is 14.7 lb/in2. What is the pressure when expressed in g/m2? (2.54 cm = 1 in

miss Akunina [59]

<h2>Answer:</h2>

14.7 lb / in² = 10333018.166 g / m²

<h2>Explanation:</h2>

Given from the question;

pressure = 14.7lb/in²

<u>To convert from lb / in² to g / m²</u>

Note;

2.54 cm = 1 in ------------(i)

But;

100 cm = 1m

=> 2.54cm = 2.54 cm x 1 m / 100 cm = 0.0254m

=>2.54cm = 0.0254m

Substitute 2.54cm = 0.0254m into equation(i)

=> 0.0254m = 1 in

2.205 lb = 1 kg ------------------(ii)

But;

1kg = 1000g

Substitute 1kg = 1000g into equation (ii)

=> 2.205 lb = 1000g ---------------------(iii)

<em>From equation(iii);</em>

if, 2.205 lb = 1000g

then, 1 lb = 1 lb x 1000 g / 2.205 lb = 453.5g

=> 1 lb = 453.5 g

<em>Now let's convert 14.7 lb/in² to g / m²;</em>

=> 14.7 lb / in² could be written as 14.7 x 1 lb / (1 in x 1 in)

i.e

=> 14.7 lb / in² = 14.7 x 1 lb / (1 in x 1 in) ------------------(iv)

<em>Substitute the values of 1 lb = 453.5g and 1 in = 0.0254m into equation(iv)</em>

=> 14.7 lb / in² = 14.7 x 453.5g / (0.0254m x 0.0254m)

=> 14.7 lb / in² = 6666.45 g / (0.00064516m²)

=> 14.7 lb / in² = 10333018.166 g / m²

<em>Therefore, 14.7 lb / in² = 10333018.166 g / m²</em>

5 0

4 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

4 years ago

An 8000-kg engine pulls a 40,000 kg train along a level track and gives it an acceleration a1 = 1.20 m/s 2 . What acceleration (

mel-nik [20]

a = F/(m1+m2)
1.2 = F/48,000
F = 57,600N

a = F/(m1+m3)
a = 57,600/24,000
a = 2.4 m/s^2

8 0

4 years ago

De onde veio o dinheiro para financiar a indústria no Brasil a partir de 1930?

liq [111]

Answer: Descomponiendo los ingresos de Brasil, encontramos que se derivan de los siguientes tres sectores: agricultura, industria y servicios. Según estimaciones de 2014, el 5,8% de los ingresos de Brasil provino de la agricultura, el 23,8% de la industria y el 70,4% de los servicios.Durante las décadas de 1940 y 1950, el gobierno brasileño comenzó a construir acerías, refinerías de petróleo y represas hidroeléctricas. ... Brasil se encuentra entre las principales naciones industriales del mundo, debido al gasohol.

Explanation:

4 0

3 years ago