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gregori [183]
1 year ago
7

Difference between insulin resistance and insulin sensitivity? with source and thank ​

Chemistry
1 answer:
Stolb23 [73]1 year ago
5 0

Answer:

Insulin resistance and insulin sensitivity are two sides of the same coin. If you have insulin resistance, you have low insulin sensitivity. Conversely, if you are sensitive to insulin, you have low insulin resistance.

Explanation:

searched it hope it helps

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Write a balenced chemical equation for the reaction: include abbreviation for the physical states-
WITCHER [35]

Answer:

MgBr(aq) + (NH4)3PO4(aq) -------> NH4Br(aq) + Mg3(PO4)2(s)

Explanation:

4 0
3 years ago
A tank at is filled with of chlorine pentafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal
Ivan

Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

= 44.53 kPa

Total Pressure exerted by the gases = 60.58 kPa

Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

Mass = 13.3 g

Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

This is Dalton's law of Partial Pressure.

- Mole fraction of Chlorine Pentafluoride

= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

3 0
2 years ago
What is the normal boiling point of etanoic acid?​
Nezavi [6.7K]

Answer: 244.6°F

Explanation:

3 0
3 years ago
When trying to determine whether or not an atom gives up electrons easily, do chemists look at electronegativity or ionization p
Archy [21]
Here I found some info at Yahoo answers: https://answers.yahoo.com/question/index?qid=20090119191941AAB7oAb
The more electronegative an atom is the more unwilling it is to lose its electrons in a compound. If you do try to take a very EN atom away from a compound you'll need to apply a lot of energy for that to happen. I can give an example of a single atom though 

<span>Cl has 7 valence electron filled and every atom wants to be like nobles (noble gases), so it's not going to give an electron away b/c it's really close to being like a noble gas. Noble gases are the most stable atoms, which is why I say stability counts.</span>
4 0
3 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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