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3 years ago

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Chemistry

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Sonja [21]3 years ago

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Organ system, because there are multiple organs that work together to help make the body function.

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Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h

Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

$k=\frac{0.693}{t_1_/_2}$

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

$k=\frac{0.693}{1.5 hrs}$

The expression for the rate constant is :

$k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}$

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

$\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6$

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

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3 years ago

The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Brrunno [24]

Answer:

The molecular formula of the compound is $C_{8}H_{16}O_{4}$ . The molecular formula is obtained by the following expression shown below

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is $C_{2}H_{4}O$

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = $\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}$

$n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4$

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Molecular formula = 4 $\times$ $C_{2}H_{4}O$

Molecular formula is $C_{8}H_{16}O_{4}$

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