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meriva
3 years ago
9

Ethyl alcohol (ch3ch2oh) is/is not soluble in water. 1. is; all organic molecules are soluble in water. 2. is; ethyl alcohol exh

ibits dipole-dipole and h-bonding interactions with water. 3. is not; ethyl alcohol exhibits dispersion forces which are insufficient for dissolving. 4. is not; ethyl alcohol is an ionic compound with a hydroxide group and hydroxides are insoluble.
Chemistry
1 answer:
nata0808 [166]3 years ago
5 0
Answer:
            Ethyl alcohol is soluble in water because <span>ethyl alcohol exhibits dipole-dipole and h-bonding interactions with water.

Explanation:
                   Ethyl alcohol and water are miscible in each other because both are polar in nature and "Like dissolves Like".
                   The bond between oxygen and hydrogen atoms, both in alcohol and water are polar in nature and results in intermolecular hydrogen bond interactions between them as hydrogen bonding results when hydrogen atom in one molecule directly attached to highly electronegative atoms like fluorine, oxygen and nitrogen forms interaction with higly electronegative atom of neighbor atom.</span>
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Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon
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In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

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3 0
3 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

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3 years ago
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The answer is 19.9 grams cadmium. 
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Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
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     mcadmium = 19.9 grams
3 0
2 years ago
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