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3 years ago

The solubility of glucose at 30°C is

Chemistry

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL \times \frac{0.996g}{1mL} =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

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What does 3.5 degree of unsaturation tell you?

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Answer:

It means the chemical entity is a radical

Explanation:

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While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.

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The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.

Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.

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3 years ago

In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a

katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

$\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole$

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

$\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}$

Now put all the given values in this formula, we get:

$0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}$

$\text{Molar mass of an acid}=266.985g/mole$

Therefore, the molar mass of an acid is 266.985 g/mole

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