Chemistry, 13.10.2019 08:30 hockeykid7583

Awhite solid can be purified from lemon juice. this solid tastes sour. when placed into water the solution conducts electricity and has a ph of 3.0. the solution turns litmus paper red. when the white powder is reacted with sodium hydroxide, a different white powder is formed which tastes salty. what kind of substance is the original white powder isolated from lemon juice?

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Answer from: svarner2001

I believe that the original white powder isolated from lemon juice is an acid. Some characteristics of an acid is being described here like tasting sour, conducts electricity when in water and having a pH less than 7. Also, turning the litmus paper into red indicates that it is an acid. Hope this answers the question. Have a nice day.

Answer from: P4thing

Explanation:

When a solution is acidic in nature then it means its pH range is from 1 to 6.9. And, in an acidic solution there are more number of hydrogen ( $H^{+}$ ) or hydronium ( $H_{3}O^{+}$ ) ions as compared to hydroxide ions ( $OH^{-}$ ).

Some properties of acids are as follows.

Acids taste sour.Acids show a pH value less that 7.Acids release hydrogen ion upon dissociation in water of polar solvents.Acids turn blue litmus into red.

Now, since the white solid substance isolated from lemon juice shows all the properties of being an acid.

Thus, we can conclude that the original white powder isolated from lemon juice is an acidic substance.

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

Answer from: Quest

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