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igomit [66]
3 years ago
9

In systemic tissue fluids, the enzyme carbonic anhydrase catalyzes the reaction co2 + h2o → h2co3 (which then can dissociate int

o h+ and hco3–). in fact, co2 released from cells is converted to hco3– ions and travels in that form in the bloodstream. hco3– is reconverted to co2 + h2o in the pulmonary capillaries by the same enzyme, and there the co2 is exhaled. how is it possible for the same enzyme to catalyze reverse reactions?
Chemistry
1 answer:
NeTakaya3 years ago
7 0
It is achievable for the same enzyme to catalyze reverse reactions for the reason that the direction of a reversible reaction is determined by the concentrations of reactants and products. In pulmonary circulation, the low CO2 concentration supports the making of CO2 and H2O.
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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
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Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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