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Olin [163]
2 years ago
14

Increased total serum lactic dehydrogenase (LD) activity, confined to fractions 4 and 5 is most likely to be associated with: A.

Pulmonary infarction B. Hemolytic anemia C. Myocardial infarction D. Acute viral hepatitis​
Chemistry
1 answer:
Ivenika [448]2 years ago
4 0
<h3>Answer:</h3>

D.Acute Viral Hepatitis

inflammation of the liver caused by infection with one of the five hepatitis viruses

Hope it helps (≧∇≦)/

You might be interested in
A chemistry student is given 250.0 mL of clear aqueous solution at 42 C. He is told an unknown amount of a certain compound X is
lara [203]

Answer:

No

Explanation:

The solubility of a solid in water refers to the amount of that solid that dissolves in water.

It is not possible to calculate the solubility of the solid because the student threw away the first precipitate that formed. We already have the volume of water, but having lost some mass of precipitate, it has become impossible to accurately determine the solubility.

Hence the answer provided above.

7 0
3 years ago
Why is creativity important in constructing scientific explanations?
poizon [28]

D. Sometimes it is necessary to put forth new theories to explain  experimental results,

Explanation:

Creativity is important to science because sometimes, it is necessary to put forth new theories to explain experimental results.

Theories are explanation of scientific observations.

  • Most scientists are critical and creative thinkers.
  • They must be able to come up with a wide range of possible explanation to explain their findings.
  • The explanations must be scientific in nature and not fall out of the basic tenets of this way of thinking.
  • Therefore, scientists draw from their wealth of deductive reasoning to give explanations into an observed phenomenon.

learn more:

Purpose of conducting experiments brainly.com/question/5096428

#learnwithBrainly

6 0
3 years ago
HELP ME PLEASE!!!!
KonstantinChe [14]
The correct answer is a.
5 0
3 years ago
Read 2 more answers
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0
3 years ago
Read 2 more answers
Please help me find the final answer
dlinn [17]

Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693.\overline{3168} grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃)  is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}

\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g

The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}

\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693.\overline{3168} grams

6 0
3 years ago
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