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Serjik [45]
2 years ago
12

A patient needs 40.0 mg of antibiotic per kilogram of body weight each day. If the patient weighs 55 kilograms.

Chemistry
1 answer:
White raven [17]2 years ago
4 0

2200 mg of antibiotic

Explanation:

Given that 40 mg of antibiotic/kg of the bodyweight is given.

If patient is 55 kg then  the dose of antibiotic will be

if 40/1000000 is done then we can get antibiotic in kg/kg of the weight

= 0.00004 kg of antibiotic per kg

0.00004*55 ( to know how much 55 kg person will require)

= 0.0022 kg

This 0.0022 value will be converted to mg

0.0022*10^6

= 2200 mg of antibiotic will be given to a 55kg patient.

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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
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Answer : The value of K_{eq} is, 11.2

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The relation between the equilibrium constant and standard Gibbs free energy is:

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where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

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T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

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Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
If 5.61 g of ammonium nitrate (NH4NO3) is dissolved in 155 mL of water to make an instant cold pack, what is the molarity of the
kifflom [539]

Answer:

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Plug in and divide.

0.0701/0.155

=0.452 M

7 0
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