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Murljashka [212]
3 years ago
13

8. What is diabetes?

Chemistry
2 answers:
DerKrebs [107]3 years ago
7 0

Answer:

Diabetes is the condition in which the body does not properly process food for use as energy. Most of the food we eat is turned into glucose, or sugar, for our bodies to use for energy. The pancreas, an organ that lies near the stomach, makes a hormone called insulin to help glucose get into the cells of our bodies.

- CDC

Ghella [55]3 years ago
6 0

Answer:

Diabetes is a disease that occurs when your blood glucose, also called blood sugar, is too high. Blood glucose is your main source of energy and comes from the food you eat. Insulin, a hormone made by the pancreas, helps glucose from food get into your cells to be used for energy. Sometimes your body doesn’t make enough—or any—insulin or doesn’t use insulin well. Glucose then stays in your blood and doesn’t reach your cells.

Over time, having too much glucose in your blood can cause health problems. Although diabetes has no cure, you can take steps to manage your diabetes and stay healthy.

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Scientists who possess this attitude always report their observations and results truthfully.
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Answer:

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3 years ago
How would a solute affect the boiling point of water?
Alex787 [66]
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Read 2 more answers
For this heterogeneous system 2 A ( aq ) + 3 B ( g ) + C ( l ) − ⇀ ↽ − 2 D ( s ) + 3 E ( g ) the concentrations and pressures at
maw [93]

Answer:

2.55*10^{11

Explanation:

Equation for the heterogeneous system is given as:

2A_{(aq)}  + 3 B_{(g)} + C_{(l)}      ⇄      2D_{(s)}    +     3E_{(g)}

The concentrations and pressures at equilibrium  are:

[A] = 9.68*10^{-2}M

P_B = 9.54*10^3Pa

[C]=14.64M

[D]=10.11M

P_E=9.56*10^4torr

If we convert both pressure into bar; we have:

P_B = 9.54*10^3Pa

P_B = (9.54*10^3)*\frac{1}{10^5} bar

P_B=9.54*10^{-2}bar

P_E=9.56*10^4torr

1 torr = 0.001333 bar

9.54*10^4 *0.001333 = 127.5 bar

K=\frac{[P_E]^3}{[A]^2[P_B]^3}

K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}

K=2.55*10^{11

3 0
3 years ago
Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---&gt; C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation
nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ...[1]

C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ..[2]

H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ..[3]

2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}

\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)

\Delta H^o_{4}=232 kJ

The enthalpy for given reaction is 232 kilo Joules.

5 0
3 years ago
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