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amid [387]
3 years ago
11

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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Answer:

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When going from top to bottom in any group of the periodic table, the atomic radius always tends to increase

8 0
2 years ago
Oxygen gas generated in an experiment is collected at 25.0°C in a bottle inverted in a trough of water. The total pressure is 1.
Nitella [24]

Answer:

0.007 mol

Explanation:

We can solve this problem using the ideal gas law:

PV = nRT

where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.

Keep in mind that when  we collect a gas over water we have to correct for the vapor pressure of water at  the temperature in the experiment.

Ptotal = PH₂O + PO₂  ⇒ PO₂ = Ptotal - PH₂O

Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.

P H₂O = 23.8 mmHg x 1 atm/760 mmHg =  0.031 atm

V = 1750 mL x 1 L/ 1000 mL = 0.175 L

T = (25 + 273) K = 298 K

PO₂ = 1 atm - 0.031 atm = 0.969 atm

n =  PV/RT = 0.969 atm x  0.1750 L / (0.08205 Latm/Kmol x 298 K)

n = 0.007 mol

6 0
2 years ago
Prove that circle p and circle q are similar.
abruzzese [7]
All circles are similar.
8 0
3 years ago
The law of universal gravitation describes the gravitational force between two objects.
Sliva [168]

Answer:

mass and distance

Explanation:

The strength of this force depends on the <u>mass</u>

of each object and

the <u>distance</u>

between them.

6 0
1 year ago
If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con
drek231 [11]

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

6 0
3 years ago
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