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ikadub [295]
3 years ago
11

If water were a linear (not bent) molecule like co2, electrostatic interactions between water molecules would be much weaker. wh

y?
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer;

The partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.

Explanation;

Water is a polar molecule because the electrons are not shared equally, they're closer to the oxygen atom than the hydrogen.

-Normally, the water molecule is a bent shape because of the pair of lone electrons - they repulse each other and exert a compression to the hydrogen atoms at a slight 104º angle. It is a bent molecular geometry that results from tetrahedral electron pair geometry.

-The 2 lone electron pairs exerts a little extra repulsion on the two bonding hydrogen atoms to create a slight compression to a 104 degrees bond angle. Therefore, the water molecule is bent molecular geometry because the lone electron pairs.

Thus, If water were a linear molecule like co2, electrostatic interactions between water molecules would be much weaker, then the partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.

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If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a
strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get:

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

3 0
3 years ago
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A student determines that 23.1 J of heat are required to raise the temperature of 6.67 g of an

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