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Keith_Richards [23]
2 years ago
14

What is the mode of action of most antifungal drugs?

Chemistry
1 answer:
Goshia [24]2 years ago
3 0

Inhibition of protein synthesis is the mode of action done by most antifungal drugs.

<h3>What is the mode of action of antifungal drugs?</h3>

Antifungals drugs inhibit the synthesis of ergosterol the main fungal sterol, polyenes which interact with fungal membrane sterols and 5-fluorocytosine inhibits macromolecular synthesis.

So we can conclude that inhibition of protein synthesis is the mode of action done by most antifungal drugs.

Learn more about drug here: brainly.com/question/26254731

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Sulfur is an element that can be seen on the periodic table. What will most likely occur if sulfur forms an ionic bond with anot
Yakvenalex [24]

What is likely  to occur  if sulfur  forms an ionic  bond with another  element  is  <u>sulfur  will accept  electrons</u>

     

    <u><em> explanation</em></u>

  • Ionic bond is formed when  a metal react with a non metal.
  • Metal  loses ( donate) electrons  to form cation ( a positively charged  ion) , while  non metal accept (gain)  electrons  to form anion ( a negatively charged ion ).

  • Sulfur  is a non metal  therefore it  accept  electrons if it form an ionic bond  with a metal. sulfur accept  2 electrons to  form S2- ion

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3 years ago
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How are they related "atom" "molecule"
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Answer:

mixture of atoms forms molecule

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2. What do acidic solutions have high concentrations of? ​
Xelga [282]
Hydrogen ions
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Which soup would likely be more flavorful, a nicely heated but not too hot to eat bowl of soup or a bowl of leftover soup direct
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Answer:

The soup that is not too hot to eat would be better

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because the flavours would be fresher than if it is leftover

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Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
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