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Kitty [74]
3 years ago
6

Calculate the total number of days therapy available within a 300 ml bottle of ranitidine (as hydrochloride) 75 mg/5 ml. oral so

lution, when it is prescribed at a dose of 300 mg at night?
Chemistry
1 answer:
Lisa [10]3 years ago
8 0

<u>Answer:</u> This therapy is available for 15 days.

<u>Explanation:</u>

We are given:

Oral solution dosage = 75 mg/5 mL

To calculate the volume of oral situation for single dose per, we use unitary method:

The volume required for 75 mg of solution is 5 mL

So, the volume required for 300 mg of solution will be = \frac{5mL}{75mg}\times 300mg=20mL

The total volume of the ranitidine bottle = 300 mL

To calculate the number of days, we divide the total volume of the bottle by the volume of dose taken per night, we get:

\text{Number of days}=\frac{\text{Total volume}}{\text{Volume of dose taken per night}}=\frac{300mL}{20mL}=15

Hence, this therapy is available for 15 days.

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How many molecules of carbon dioxide are in 5.61 moles of carbon dioxide (CO2)?
gulaghasi [49]
<h3>Answer:</h3>

3.38 × 10²⁴ molecules CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 5.61 moles CO₂

[Solve] molecules CO₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 5.61 \ mooles \ CO_2(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 3.37834 \cdot 10^{24} \ molecules \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂

6 0
3 years ago
What is the volume of 500g of CO2?
Airida [17]

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter


Try to see if this helps

3 0
3 years ago
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What is a zone of weak, variable winds located at 30 degrees north and 30 degrees south?
timurjin [86]

Answer:

Horse latitude, trade winds

Explanation:

  • The area of the low pressure or the calm consists of the variable light winds that blow near the equator are known to the marines as the doldrums and they form a circuital pattern near the earth atmosphere.
  • Forms at a center of the near the higher pressure systems called as the horse latitudes where the  trade winds at the surface are weak and variable and this zone is found generally in latitudes of the 30° North and South of the equator and move in an east to west direction.

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3 years ago
What does the model represent
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The model represents Photosynthesis. Light, carbon dioxide, and water are the reactants. The products will be Sugar and Oxygen. Photosynthesis releases oxygen and glucose. Which is the energy that living things need to survive.
6 0
3 years ago
The normal boiling point of methanol is 64.7 ∘ C and the molar enthalpy of vaporization is 71.8kJ/mol . The value of ΔS when 1.3
guapka [62]

Answer:

286 J/K

Explanation:

The molar Gibbs free energy for the vaporization (ΔGvap) is:

ΔGvap = ΔHvap - T.ΔSvap

where,

ΔHvap: molar enthalpy of vaporization

T: absolute temperature

ΔSvap: molar entropy of the vaporization

When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.

ΔHvap - Tb . ΔSvap = 0

ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol

When 1.35 mol of methanol vaporizes, the change in the entropy is:

1.35mol.\frac{212J}{K.mol} =286 J/K

7 0
3 years ago
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