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topjm [15]
3 years ago
10

Which type of hybridization leads to a bent molecular geometry and a tetrahedral electron domain geometry? sp s2p s2p2 sp3

Chemistry
2 answers:
Setler [38]3 years ago
7 0
Your answer is going to be sp3. :)
Orlov [11]3 years ago
3 0

Answer is: sp3 hybridization.

Water (H₂O), oxygen difluoride (OF₂) and sulfur(IV) oxide (SO₂) are examples of bent molecules.

Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.

Water is polar because of the bent shape of the molecule.

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A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
A naturally occurring element consists of three isotopes. The data for the isotopes are:
max2010maxim [7]
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</span>= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u
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Answer:

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Explanation:

ols the. atom's image didn't appear

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Answer: c i think

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