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dusya [7]
2 years ago
9

Bromine is less electronegative than chlorine, yet methyl bromide and methyl chloride have very similar dipole moments. why? sel

ect the single best answer.
Chemistry
1 answer:
diamong [38]2 years ago
4 0
Bromine is less electronegative than chlorine, yet methyl bromide and methyl chloride have very similar dipole moments. This is because the bond distance in methyl bromide is more due to the large size of bromine atom.
Dipole moment is calculated by multiplying the charge on the atom with the bond distance.
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0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33g of lead. Determine the simplest formula for the compound.
Valentin [98]
PbO2

You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula  would be PbO2

8 0
2 years ago
Consider the two facts below:
OLEGan [10]

Answer:

A. There is more dissolved oxygen in colder waters than in warm water.

D. If ocean temperature rise, then the risk to the fish population increases.

Explanation:

Conclusion that can be drawn from the two facts stated above:

*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.

*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.

*Fishes, therefore, would thrive best in colder waters than warmer waters.

The following are scenarios that can be explained by the facts given and conclusions arrived:

A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)

D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).

4 0
1 year ago
Draw a second resonance structure for the following ion (be sure to include the charges and all lone pairs)... + ..N=N=N <---
joja [24]

Answer:

Explanation:

Resonance structure occurs in an organic compound that undergoes resonance effects. This resonance effect is sometimes called the mesomeric effect helps to increases the stability of organic compounds that have alternating single bonds and double bonds.

The second resonance structure diagram for the ion given in the question can be found in the attached diagram below.

8 0
1 year ago
6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ
qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy (\Delta G) should be negative.

We know, \Delta G=\Delta H-T\Delta S, where T is temperature in Kelvin scale.

Reaction A: \Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{10.5\times 10^{3}}{30}

or, T> 350

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: \Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{1.8\times 10^{3}}{-113}

or, T> -16

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0
2 years ago
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