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anzhelika [568]
3 years ago
6

Analyses of enzymes found in the blood are used as indicators that tissue damage (heart, liver, muscle) has occurred and resulte

d in the leakage of cellular enzymes into the bloodstream.
a. True
b. False
Chemistry
2 answers:
Reika [66]3 years ago
7 0

Answer:

True.

Explanation:

Yes, analyses of enzymes found in the blood are used as indicators of tissue damage in the heart, liver, muscle etc has occurred. This leakage of enzymes into the bloodstream tells us whether the tissue is damaged or not. Lactate dehydrogenase is a type of enzyme which is used as indicator which is responsible for the interconverts lactate and pyruvate. The concentration of this enzyme in the blood tells us about tissue damage.

Lunna [17]3 years ago
7 0

Answer:

a. True

Explanation:

The co enzyme used for this reaction is NADH+, here H+ acts as co substrate.

This is required to find the tissue damage that has occurred in our body which resulted in leakage of cellular enzymes into the bloodstream.

This coenzyme is used for conversion of lactose dehydrogenase.

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An unknown insoluble substance displaced the water shown. It's mass is indicated on the triple beam balance. Mass = A. 694 B. 69
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Answer:

D. 693.8.

Explanation:

  • The first scale of the triple beam balance indicate that the mass is 600.
  • The second scale of the triple beam balance indicate that the mass is 690.
  • The third scale of the triple beam balance indicate that the mass is higher that 693.5 and the scale is near but lower than 694, so the mass = 693.8.c
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How many seconds are in 5 hours
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Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What
Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For CH_4  :-

Mass of CH_4  = 1.28 g

Molar mass of CH_4  = 16.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.28\ g}{16.04\ g/mol}

Moles_{CH_4}= 0.0798\ mol

For O_2  :-

Mass of O_2  = 10.1 g

Molar mass of O_2  = 31.998 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{10.1\ g}{31.998\ g/mol}

Moles_{O_2}= 0.3156\ mol

According to the given reaction:

CH_4+2O_2\rightarrow CO_2+2H_2O

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, CH_4 is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0798\ moles= \frac{Mass}{44.01\ g/mol}

Mass of CO_2 = 3.51 g

<u> Theoretical yield = 3.51 g</u>

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