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3 years ago

What is a result of the unequal electron sharing in a water molecule?

Chemistry

1 answer:

Katarina [22]3 years ago

4 0

Answer:

The right choice is c. Water molecules have a weakly positive hydrogen end.

Explanation:

The unequal sharing of electrons in water molecule gives a slight negative charge near its oxygen atom ( see image below) and a slight positive charge near its hydrogen atoms. A neutral molecule that has a partial positive charge at one end and a partial negative charge at the other, it is a polar molecule.

a. Water molecules have a nonpolar bond.

It is wrong choice because water has polar bond .

b. Water molecules have a weakly positive oxygen end.

Also, a wrong choice due to water molecule gives a slight negative charge near its oxygen atom.

c. Water molecules have a weakly positive hydrogen end.

This is the right choice.

d. Water molecules have two oxygen and two hydrogen atoms

It is wrong choice because water has one oxygen and two hydrogen atoms

<u>So, the right choice is</u>

c. Water molecules have a weakly positive hydrogen end.

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Writing Chemical Equations

nignag [31]

Answer:

The prefix di means 2 and the prefix hexa means 6.

As per IUPAC rules you must use the prefixes to indicate the number of atoms of each element in the formula.

So, the proper name for the compound C2H6 is dicarbon hexahydride.

Explanation:

8 0

2 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0

3 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

Based on the Lewis/electron dot representation

timama [110]

Answer:

B or C think so

4 0

2 years ago

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to

Rainbow [258]

Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: $\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }$

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

$56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm$

Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.

$P_{2} =\frac{T_{2} P_{1} }{T_{1} }$

$P_{2}=\frac{(337K)(0.55atm)}{418K}$

P₂=0.44 atm

8 0

3 years ago