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KIM [24]
3 years ago
6

A glucose solution has a density of 1.02 g/ml. what is its specific gravity

Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
7 0

<u>Answer:</u> The specific gravity of glucose solution is 1.02

<u>Explanation:</u>

The relationship between specific gravity and density of a substance is given as:

\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}

We are given:

Density of glucose solution = 1.02 g/mL

Density of water = 1.00 g/mL

Putting values in above equation we get:

\text{Specific gravity of glucose solution}=\frac{1.02g/mL}{1.00g/mL}\\\\\text{Specific gravity of glucose solution}=1.02

Hence, the specific gravity of glucose solution is 1.02

Makovka662 [10]3 years ago
4 0
Answer is: specific gravity of glucose is 1,02.
d(glucose) = 1,02 g/ml.
d(water) = 1,00 g/ml.
Specific gravity of glucose = density of glucose ÷ density of water.
Specific gravity of glucose = 1,02 g/ml ÷ 1,00 g/ml.
Specific gravity of glucose = 1,02.
Specific gravity<span> is the ratio of the </span>density<span> of a substance (in this case glucose) to the density of a reference substance (water).</span>
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7 0
2 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
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3 years ago
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