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Evgen [1.6K]
1 year ago
15

water interacts with polar substances like –oh groups but not with non-polar substances like methyl (−ch3) or ethyl groups (−ch2

). using this information, select the molecules that will easily dissolve in water based on polarity not on size. check all that apply.
Chemistry
1 answer:
Brilliant_brown [7]1 year ago
7 0

CH3OH and CH3CH2CH2OH will easily dissolve in water based on polarity not on size.

<h3>What is Polarity ?</h3>

In chemistry, polarity describes the type of bonds that exist between atoms. Atoms share electrons when they join forces to form chemical bonds. When one of the atoms applies a stronger attractive force to the bond's electrons, a polar molecule is created.

  • For instance, the chlorine atom is slightly negatively charged while the hydrogen atom in hydrogen chloride is slightly positively charged.

  • Water is polar due to its form even though its molecules have no net charge. The molecule's hydrogen ends are positive, and its oxygen ends are negative. Water molecules are drawn to other polar molecules and to one another as a result.

Learn more about Polarity here:

brainly.com/question/26849258

#SPJ4

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

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

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

3 0
3 years ago
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