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3 years ago

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How many kilograms of oseltamivir would be needed to treat all the people in a city with a population of 400000 people if each p

erson consumes two oseltamivir capsules a day for 5 days? Express your answer to one significant figures and include the appropriate units.

1 answer:

enyata [817]3 years ago

6 0

Answer:

300 kg

Explanation:

The give parameters obtained from a similar question are;

The mass of oseltamivir per capsule = 75 mg

The mass of oseltamivir required, <em>M</em>, is given as follows;

M = The mass of oseltamivir per capsule × The mass taken per person per day × 5 days × The number of people in the city

M = 0.75 mg/capsule × 2 capsule/(day·person) × 5 days × 400,000 people

M = 300,000,000 mg = 300 kg

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You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the

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Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

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And the resulitng moles of HX = Initial moles - Moles NaOH added

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Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

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inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

$Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)$

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

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