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irinina [24]
3 years ago
7

A formula for a cough syrup contains 60 mg of codeine per fluid ounce. How many mg are contained in one teaspoonful?

Chemistry
1 answer:
Kruka [31]3 years ago
7 0

Answer : The amount of codeine in one teaspoonful is, 30 mg

Explanation : Given,

Amount of codeine per fluid ounce = 60 mg

Now we have to determine the amount of codeine present in one teaspoonful.

As we know that:

1 teaspoonful = 0.5 fluid ounce

As, the amount of codeine per fluid ounce = 60 mg

So, the amount of codeine 0.5 fluid ounce = \frac{0.5}{1}\times 60mg=30mg

Thus, the amount of codeine in one teaspoonful is, 30 mg

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Use the atomic mass of indium to calculate the relative abundance of indium-113.
ASHA 777 [7]

The relative abundance of indium-113 is 4%.

The isotopes are species of the same element having the same atomic number but a different mass number.

The elements occurring in nature exist as multiple isotopes.

When we take into account the existence of these isotopes and their relative abundance (percent), the average atomic mass of that element can be computed, which is given by the following formula,

Average atomic Mass= (%age of isotope 1) x (Mass of isotope 1) + (%age of isotope 2) x (Mass of isotope 2)/100

Indium exists in the form of Indium-113 and Indium-115.

The mass of Indium-113 is 112.90 u.

The mass of Indium-115 is 114.90 u.

The average atomic mass of Indium is 114.82 u.

Let the %age of isotope 1(Indium-113) be X.

Then, the %age of isotope 2(Indium-115) would be 100-X.

Applying the values in the formula,

Average atomic mass = 112.90X+114.90(100-X)

114.82 = 112.90X+114.90(100-X)

On solving the above equation, the value of X comes out to be 4%.

Thus, the relative abundance/%age abundance of Indium-113 is 4%.

To know more about "Average Atomic Mass", refer to the following link:

brainly.com/question/13753702?referrer=searchResults

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7 0
1 year ago
What happens to the energy that is lost when water freezes?
SVEN [57.7K]
<span>As we know through the principle of conservation of energy, energy can neither be created nor destroyed. Therefore, the energy removed from the water in order to make it freeze is absorbed by the surroundings. This is why the surroundings in which freezing is taking place are below freezing. This is more easily illustrated in the example of condensation. If you were to hold a plate over a pot of boiling water, some of the water would give its energy to the plate and condense on its surface.</span>
4 0
3 years ago
Read 2 more answers
How many grams of FeCl 3 are in 250. mL of a 0.100 M solution?
Naddika [18.5K]

Answer:

The correct answer is option B

Explanation:

$Molarity=\frac{Weight}{Molecular \,weight} \frac{1000}{V(in \, ml)}

Given values,

Molarity of $FeCl_3=0.100M$

Volume of solution, $V=250ml$

Molecular weight of $FeCl_3=162.2$

Substituting this values in Molarity formula, we get

$0.1=\frac{weight}{162.2} \times\frac{1000}{250} $\\$\Rightarrow 16.22=weight\times4$\\$\Rightarrow weight=\frac{16.22}{4} $\\$\therefore weight=4.06g$

5 0
3 years ago
Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?.
Julli [10]

HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>
  • The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
  • Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
  • For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
  • When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
  • A buffer made of a weak acid and its salt is an example.
  • It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.

learn more about buffer solutions here

<u>brainly.com/question/8676275</u>

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8 0
1 year ago
A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume
jenyasd209 [6]
Volume of Argon V1 = 5.0 L  
Pressure of Argon P1 = 2 atm 
Final temperature T2 = 30 C = 30 + 273 = 303 K 
Volume at final temperature V2= 6 L 
Pressure at final temperature P2 = 8 atm  
We know that (P1 x V1) / T1 = (P2 x V2) / T2  
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48 
Initial Temperature T1 = 3030 / 48 = 63.12 
Initial Temperature = -209. 8 C
4 0
3 years ago
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