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Murrr4er [49]
3 years ago
15

In a titration, what is the name of the substance that is being determined?

Chemistry
1 answer:
solmaris [256]3 years ago
8 0
<span>In a titration, the substance that is unknown and being identified is called analyte. A titration is where a known solution or concentration called the titrant is used to identify and measure an unknown substance which is the analyte.</span>
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7 0
3 years ago
Given the equation representing a reversible reaction:
tester [92]

Answer:

- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)

Explanation:

Hello,

In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:

CH_3COOH(aq) + H_2O \rightarrow CH_3COO^-(aq) + H_3O^+(aq)

It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:

CH_3COO^-(aq) + H_3O^+(aq)\rightarrow CH_3COOH(aq) + H_2O

It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.

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5 0
3 years ago
Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

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6 0
3 years ago
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