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3 years ago

A patient needs 0.024 g of a sulfa drug. There are 12-mg tablets in stock. How many tablets should be given?

2 answers:

7 0

I would personally convert the 12 mg to g so I could see what I was working with. So 12 mg to grams is 0.012 g...

so 1 tablet is 0.012g. the patient needs 0.024 g.

so 0.024g/0.012g = 2 tablets or 0.012g X 2 is 0.024 g

hope this helps :)

enyata [817]3 years ago

6 0

Answer : The number of tablets given should be, 0.5

Explanation :

We are given :

Mass of sulfa drug for a patient = 0.024 g

Total mass of tablets = 12 mg = $12\times 10^{-3}g=0.012g$

(Conversion factor: 1 g = 1000 mg)

To calculate the number of tablets, we divide the total mass of tablets to the mass of sulfa drug for a patient.

$n=\frac{\text{Total mass of tablets}}{\text{Mass of sulfa drug for a patient}}$

Putting values in above equation, we get:

$n=\frac{0.012g}{0.024g}\\\\n=0.5$

Hence, the number of tablets given should be, 0.5

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Certain gases in the atmosphere trap heat on Earth.

ValentinkaMS [17]

Answer:

Greenhouse

Explanation:

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3 years ago

A standard gold bar has a mass of 12.4 kg. When the gold bar is placed into a container of water the volume changes by 642 mL. W

AlekseyPX

Answer:

19.3 g/mL

Explanation:

The following data were obtained from the question:

Mass (m) = 12.4 kg

Volume (V) = 642 mL.

Density (D) =.?

Next, we shall convert 12.4 kg to grams. This can be obtained as follow:

1 kg = 1000 g

Therefore,

12.4 kg = 12.4 × 1000

12.4 kg = 12400 g

Therefore, 12.4 kg is equivalent to 12400 g.

Finally, we shall determine the density of the gold as follow:

Density is simply defined as the mass of the substance per unit volume of the substance. It can be represented mathematically as:

Density (D) = mass (m) / volume (V)

D = m/V

With the above formula, the density of gold can easily be obtained as follow:

Mass (m) = 12400 g

Volume (V) = 642 mL.

Density (D) =.?

D = m/V

D = 12400/642

D = 19.3 g/mL

Therefore, the density of hold is 19.3 g/mL

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3 years ago

The two isotopes of chlorine are 35/17 cl and 37/17cl. which isotope is the most abundant

Natalka [10]

The most abundant is 35/17 Cl

explanation

This is because Chlorine 35/17 is the one close to the average atomic mass of chlorine which is 35.46 u making it to be the most abundant.

average atomic mass of chlorine is 35.46 + or- 0.02

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3 years ago

What process takes place before an article is published in a scientific journal/

Goshia [24]

Before an article is published in a scientific journal or in any "peer-reviewed" journal the article is reviewed thoroughly by scholars from the journal as well as peers or scholars of the articles author from the same field. This process occurs to provide credibility to the ideas being published and so that readers and other scholars can rely on the validity of the material being published.

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3 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Len [333]

Answer:

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²) where λ is the wavelength

Rh is Rydberg constant

n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a. n₁ = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b. n₁ = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/4² - 1/5²) = 2.468 x 10⁵/m

λ = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x (1 x10⁹ nm/m) = 4051 nm

c. n₁ = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/5²) = 7801 x 10⁵/m

λ = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

8 0

3 years ago