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Nastasia [14]
3 years ago
7

A medicine is given 73 grains ap/lbs. of patient per dose. The patient weighs 65.7 kilograms. how many drams ap should the apoth

ecary tell the patient to take per dose?
Chemistry
1 answer:
erastova [34]3 years ago
3 0

10,573.4

Explanation:

Given parameters:

Dosage per patient = 73ap/lbs

Weight of patient = 65.7kg

Unknown:

Amount of grams/ap to be given to the patient;

 

Solution:

Since our final answer should be expressed in grams, let us convert all units into grams.

   1000g = 1kg

 Converting 65.7kg to g = 65.7 x 1000 = 65700g

73ap/lbs;

      1lb = 453.6g

 73ap/lbs x \frac{1lb}{453.6g} = 0.161ap/g

Now,

 For a 65700g person, the number of grains app required;

      65700 x 0.161 = 10,573.4grains ap per dose

learn more:

Mass brainly.com/question/3764212

#learnwithBrainly

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How many atoms of carbon are found in a 30.0 gram sample of the element?
saveliy_v [14]

Answer:

1.50 × 10²⁴ atoms C

Explanation:

Step 1: Define

Molar mass of C - 12.01 g/mol

Avagadro's Number: 6.02 × 10²³ atoms, molecules, formula units, etc.

Step 2: Use Dimensional Analysis

30.0 \hspace{3} g \hspace{3} C(\frac{1 \hspace{3} mol \hspace{3} C}{12.01 \hspace{3} g \hspace{3} C} )(\frac{6.02(10)^{23} \hspace{3} atoms \hspace{3} C}{1 \hspace{3} mol \hspace{3} C} ) = 1.50375 × 10²⁴ atoms C

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3 0
3 years ago
(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed
soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃  → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • KClO₃: 2 moles  
  • KCl: 2 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • KClO₃: 122.45 g/mole
  • KCl: 74.45 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
  • KCl: 2 moles ×74.45 g/mole= 148.9 grams
  • O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that the gas occupies.
  • T is the temperature of the gas.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

  • P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
  • V= 165 mL= 0.165 L (being 1000 mL= 1 L)
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 \frac{atmL}{molK}× 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 \frac{atmL}{molK}× 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

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