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nataly862011 [7]

3 years ago

12

Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3C

HO CH3OH CH3CN

1 answer:

RSB [31]3 years ago

3 0

Answer:

CH3CH2CH3

Explanation:

Dipole moment is the measure of the polarity of a chemical bond. It is the extent of charge separation in a molecule.

Dipole moment is the product of the magnitude of charge and the distance separating the charges from each other.

The molecule having the lowest dipole moment among the options is the molecule that has the least polarity. The least polar molecule among the options is CH3CH2CH3, it has no polar bonds in its structure.

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A beaker containing a mixture of N2(g) and H2(g) is cooled by placing it in a tray of cold water. In response to the cooling, th

Nataliya [291]

The frequency of collisions between the N₂ and H₂ molecules will decrease and the rate of reaction will also decrease.

Since the water is cooler than the gas mixture, heat will flow from the gas to the water.

The gas will cool down, so the average kinetic energy of the gas molecules will decrease.

The molecules will be moving more slowly, so there will be <em>fewer collisions</em> and <em>fewer of these collisions will have enough energy to react</em>.

The rate of reaction between H₂ and N₂ molecules at room temperature is exceedingly slow, <em>but cooling the gas mixture will make the reaction even slower</em>.

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4 years ago

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Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)

riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

$2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(Br_2)}$ = 245.463 J/mol.K

$\Delta S_f^0_{(F_2)}$ = 202.78 J/mol.K

$\Delta S_f^0_{(BrF_3)}$ = 292.53 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]$

$\Delta S^o=268.74J/K$

Now we have to calculate the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of $BrF_3$ has entropy change = 268.74 J/K

So, 1.62 moles of $BrF_3$ has entropy change = $\frac{1.62}{2}\times 268.74=217.68J/K$

Therefore, the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

3 0

4 years ago

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

Pavlova-9 [17]

Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

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

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

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3 years ago

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