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earnstyle [38]
3 years ago
5

An average human being has about 5.0 L of blood in his or her body. If an average person were to eat 37.7 g of sugar (sucrose, ,

342.30 g/mol), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change
Chemistry
1 answer:
Viktor [21]3 years ago
6 0

Answer: 0.0220275 M

Explanation:

So, we are given the following data or parameters which are going to help in solving this particular Question/problem.

=> Averagely, we have the volume = 5.0 L of blood in human body .

=> Mass of sugar eaten = 37.7 g of sugar (sucrose, 342.30 g/mol).

Therefore, the molarity of the blood sugar change can be calculated as below:

The molarity of the blood sugar change = (1/ volume) × mass/molar mass.

Thus, the molarity of the blood sugar change = (1/5) × 37.7/342.30 = 0.0220275 M.

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Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

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Explanation:

Given data:

Initial volume of gas = 5.00 L

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Final volume = 1100 mL, 280 L, 87.5 mL

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Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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