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3 years ago

Use complete sentences to differentiate between acids and bases on the basis of taste. Give an example of each type.

2 answers:

5 0

It is not advisable to differentiate acids and bases through taste because some are toxic and inedible. But for those that are safe to ingest, acid tastes sour like vinegar and lemonade. Alkaline or basic solutions taste bitter like if you'vetried tasting soap detergent as you peel it open with your teeth.

FinnZ [79.3K]3 years ago

3 0

Acid is a molecule capable of donating hydrogen ion and they form aqueous solutions with a sour taste while base is a substance that accepts proton from proton donor and in aqueous solution, they have an astringent or bitter taste. Moreover, a good example for base is sodium hydrogen carbonate as baking soda or baking powder and for acid, the most common example is the acetic acid or vinegar.

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Due Tomorrow! What is [OH-] of a solution with a pH of -1.0? Answer in M

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Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

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30 mL of 0.6 M HCl solution is neutralized with 90 mL NaOH solution. What is the concentration of the base? Show all work Please

skad [1K]

From

na=1

nb=1

Va=30mls

Vb=90mls

Ma=0.6M

Mb=?

from

Mb=Ma×Va×nb

 Vb×na

Mb=0.6×30×1

 90×1

mb=18

 90

Mb=0.2M

NaOH=40

From

conc=molarity×mr

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2 years ago