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3 years ago

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How do hydrogen and oxygen atoms form covalent bonds to create water? A: Oxygen gives its electrons to hydrogen. B: Hydrogen giv

es its electrons to oxygen. C: Oxygen and hydrogen share electrons with each other.

2 answers:

Dvinal [7]3 years ago

4 0

C) Oxygen and hydrogen share electrons with each other.

Thepotemich [5.8K]3 years ago

3 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

Covalent bond is formed when there is sharing of electrons between the atoms that are forming a bond.

When hydrogen and oxygen form a covalent bond to form a compound named water, both the atoms share electrons with each other.

Electronic configuration of Hydrogen = $1s^1$

Electronic configuration of Oxygen = $1s^22s^22p^4$

As, 2 electrons are required for one oxygen atom to complete its octet and 1 electron is required by the hydrogen to have a stable configuration. Therefore, 2 atoms of hydrogen shares its electrons to 1 atom of oxygen to form water.

The chemical formula for water is $H_2O$ .

Hence, the correct answer is option C.

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In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th

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Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

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From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

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3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

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Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

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d) $a=24.07mm/s^{2}$

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In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

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$\omega=\frac{2\pi}{86400s}$

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$a_{c}=33.73mm/s^{2}$

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In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

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$N-mg+ma_{c} = 0$

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$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

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c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

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$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

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$r= R_{E}cos \theta$

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$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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Answer:

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1 year ago

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3 years ago

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3 years ago