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Dmitry_Shevchenko [17]
3 years ago
10

Glucose is absorbed into the blood in the small intestine. Most of the glucose is absorbed by diffusion. How does the glucose co

ncentration in the blood compare to the glucose concentration in the small intestine? *
Pick one please
The concentration in the blood is higher.
The concentration in the blood is lower.
The concentration in the blood is the same.
None of the above.
Chemistry
2 answers:
Agata [3.3K]3 years ago
8 0

Answer:

Hello There!!!

Explanation:

The answer is=>The concentration in the blood is lower

Hope it helps,have a great day!!!

~Pinky~

Sonbull [250]3 years ago
4 0

Answer: The concentration in the blood is lower

Explanation:

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What element has a half-life = to 64
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The best answer is the isotope of strontium which is strontium-85. It has a half-life of about 64 days. The metal strontioum has four stable, naturally occurring isotopes which includes 84Sr (0.56%), 86Sr (9.86%), 87Sr (7.0%) and 88Sr (82.58%).
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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
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m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

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