Ask question

Ask question

All categories

3 years ago

6

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

ours for an adolescent. What fraction of the original reactant concentration will remain after 8 hours if the original concentration was 8.4 x 10-5 M.

1 answer:

Genrish500 [490]3 years ago

5 0

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.

You might be interested in

How many moles of hydrogen are in 3. 06 × 10⁻³ g of glycine , c₂h₅no₂?.

Gala2k [10]

Answer:

n = 6.06 x $10^{-4}$ mol

Explanation:

n =?

m = 3.06 x 10-³ g

M (H5) = 5 x 1.01 (Since we only want hydrogen)

Atomic mass of C = 12.01

Atomic mass of H is 1,01, etc.

Having this data, we can use the Molar mass formula and change it so we can know the quantity of matter (n) in moles, and we just replace it.

M = $\frac{m}{n}$ ⇔ n = $\frac{m}{M}$ ⇔ n = $\frac{3.06 x 10^{-3} }{5,05}$ ⇔ n = 6.06 x $10^{-4}$ mol

Note: The numbers I've used may be different from yours, by a small difference. I don't know if it's the case, but hope it helped.

8 0

2 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago

What is k in the rate law equation?

REY [17]

Answer:

B. A rate constant

Explanation:

The mathematical expression of rate law is given below,

Rate = K[A]m[B]n

This rate law show the relationship between the rate of chemical reaction and concentration of reactants.

In given equation [A] and [B] are molar concentration of reactants while K represent rate constant.

The value of K is specific for particular reaction at particular temperature,

m and n are represent exponents and determine experimentally. The value of K is not depend upon the concentrations of reactant but depend upon the surface area and temperature

5 0

3 years ago

Read 2 more answers

A sealed container at 25oC contains a gas at a pressure of 104 kPa. What is the pressure of the gas when it is heated to 225oC?

seropon [69]

Answer:

174 kPa

Explanation:

Given that,

Initial temperature, T₁ = 25° C = 25+273 = 298 K

Final temperature, T₂ = 225°C = 225 + 273 = 498 K

Initial pressure, P₁ = 104 kPa

We need to find the new pressure. The relation between the temperature and pressure is given by :

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

So,

$P_2=\dfrac{P_1T_2}{T_1}\\\\P_2=\dfrac{104\times 498}{298}\\\\P_2=173.79\ kPa$

or

P₂ = 174 kPa

So, the new pressure is 174 kPa.

4 0

3 years ago

Which type of climate does Florida have due to its latitude?

nevsk [136]

The land and the way the heat changes around there

5 0

3 years ago