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3 years ago

Why do polar and nonpolar molecules have different properties

Physics

1 answer:

kondaur [170]3 years ago

5 0

Explanation:

In case of Polar molecules electron shared are unequal that's why they have an uneven distribution of charges as whereas non-polar molecules do not the distribution of charge as the electrons are shared equally that the reason they both differ from each other.

Examples of polar molecules are water, ethanol,ammonia and sulfur dioxide.

Examples of non polar molecules are oxygen,ozone,carbon dioxide,nitrogen.

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Which color of light has the greatest frequency?

olga_2 [115]

Violet is at the end of the spectrum so there for it has the highest frequency

7 0

3 years ago

A book prone to air resistance is released from rest 300 m

yaroslaw [1]

Answer:

Approximately $73\%$ .

(Assuming that $g = \rm 9.81\; m \cdot s^{-2}$ .)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let $m(\text{book})$ represent the mass of the book.

The initial potential energy of the book is

$\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}$ .

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be $(2.943\times 10^3) \cdot m(\text{book})$ .

When the book was about to hit the ground, its speed is $\rm 40\; m \cdot s^{-1}$ . Its kinetic energy would be:

$\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}$ .

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be $\left(8.00\times 10^2\right)\cdot m(\text{book})$ when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

$\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}$ .

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, $m(\text{book})$ , was eliminated in this process.

$\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}$ .

5 0

4 years ago

Nuclear Reactions:<br> Please answer numbers 4,5,6.

Vikentia [17]

Answer:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

Explanation:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + __

Let ᵇₐX be the unknown.

Thus, the equation becomes:

¹₁H + ¹³₆C —> ¹⁴₇N + ᵇₐX

Next, we shall determine b, a and X. This is illustrated below:

For b:

1 + 13 = 14 + b

14 = 14 + b

Collect like terms

b = 14 – 14

b = 0

For a:

1 + 6 = 7 + a

7 = 7 + a

Collect like terms

a = 7 – 7

a = 0

Therefore,

ᵇₐX => ⁰₀X => γ

Thus, the balanced equation is

¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> __ + ⁴₂He

Let ˣᵧA be the unknown.

Thus, the equation becomes:

¹₀n + ¹⁰₅B —> ˣᵧA + ⁴₂He

Next, we shall determine x, y and A. This can be obtained as follow:

For x:

1 + 10 = x + 4

11 = x + 4

Collect like terms

x = 11 – 4

x = 7

For y:

0 + 5 = y + 2

5 = y + 2

Collect like terms

y = 5 – 2

y = 3

Therefore,

ˣᵧA =>⁷₃A => ⁷₃Li

Thus, the balanced equation is:

¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> __ + ¹₁H

Let ᶜₑG be the unknown.

Thus, the equation becomes:

⁴₂He + ¹⁴₇N —> ᶜₑG + ¹₁H

Next, we shall determine c, e and G. This can be obtained as follow:

For c:

4 + 14 = c + 1

18 = c + 1

Collect like terms

c = 18 – 1

c = 17

For e:

2 + 7 = e + 1

9 = e + 1

Collect like terms

e = 9 – 1

e = 8

Therefore,

ᶜₑG => ¹⁷₈G => ¹⁷₈O

Thus, the balanced equation is:

⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

8 0

3 years ago

What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?

ivann1987 [24]

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

4 0

3 years ago

Which kind of pressure prevents stars of extremely large mass from forming?

elena-14-01-66 [18.8K]

Answer:

Radiation pressure

Explanation:

Radiation pressure is gotten from the light source. The radiation pressure has a set mass limit of 150 solar masses. Any mass from 150 solar masses or below supports the formation of the stars.

However when the mass exceeds 150 solar masses the force of gravity is unable to hold the masses together and the excess mass

Is usually blown away after which the required mass is then used in the formation.

3 0

4 years ago

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