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k0ka [10]
3 years ago
10

The recommended dose for acetaminophen is 10.0 to 15.0mg/kg of body weight for adults using this guideline calculate the maximum

single dosage for a person who weighs 209lb
Chemistry
1 answer:
Oxana [17]3 years ago
4 0

Answer:

1422mg of acetaminophen

Explanation:

The maximum dose of acetaminophen is 15.0 mg of acetaminophen per kg of person.

To know the maximum single dosage of the person we need to convert the 209lb to kg (Using 1kg = 2.2046lb):

209lb * (1kg / 2.2046lb) = 94.8

The person weighs 94.8kg and the maximum single dosage for the person is:

94.8kg * (15.0mg acetaminophen / kg) =

1422mg of acetaminophen

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Which sentences describe decomposers in a food chain?
Elan Coil [88]

Answer:

hey are the final link in the energy flow in a food chain or a food web. They are fungi and animals that feed on dead organic mattet.

Explanaticuz im naruto uzamaki

5 0
3 years ago
A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan
Nookie1986 [14]

<u>Answer:</u>

<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

<u>Explanation:</u>

<u>For 1:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For helium:</u>

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol

  • <u>For methane gas:</u>

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol

To calculate the mole fraction , we use the equation:

\chi_A=\frac{n_A}{n_A+n_B}     .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}       ......(3)

  • <u>For Helium gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{He}=\frac{0.181}{0.181+0.214}=0.458

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{He}=0.458

Putting values in equation 3, we get:

p_{He}=0.458\times 821mmHg=376mmHg

  • <u>For Methane gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{CH_4}=0.542

Putting values in equation 3, we get:

p_{CH_4}=0.542\times 821mmHg=445mmHg

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

  • <u>For 2:</u>

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

<u>For nitrogen gas:</u>

363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392

<u>For carbon dioxide gas:</u>

564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0
3 years ago
If you are operating a power driven vessel that is underway in conditions of restricted visibility, what should you do
swat32
If you are operating a power driven vessel that is underway in condition of restricted visibility, you are expected to do the following: sound prolonged blasts every two minutes. If the vessel is underway but is not moving, it is expected to sound two prolonged blast every two minutes. When one hears any of the signal above, one is expected to reduce speed to the minimum that is needed to keep on course.
3 0
3 years ago
You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t
Anna [14]

Answer:

  • <em>The mystery substance is</em> <u>C. Bromine (Br) </u>

Explanation:

<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, ​and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts  producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, <em>the mystery substance is bromine (Br).</em>

7 0
3 years ago
What is an intermolecular force?
viva [34]

Answer: a force acting between two different molecules

Explanation:

APEX

4 0
3 years ago
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