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2 years ago

If a solution contains 20g of glycerol and 5g of coeine, what is the codeine percentage strength of the solution?

Chemistry

1 answer:

Strike441 [17]2 years ago

5 0

20%

The codeine percentage strength of the solution is 20%

The term "percent strength" describes the proportion of a material that has dissolved in a given volume of liquid.
Knowing how parts relate to wholes is the secret of % strength: A percent is equal to x parts out of 100 overall.

<h3>Meaning of percentage strength</h3>

The amount of a liquid is often measured, and the % strength tells you how many parts by volume of the active ingredient are present.
100 parts by volume are included in the total volume of the solution or liquid preparation.

<h3>How do you determine a solution's strength?</h3>

Mass of solute times strength of solution(g) Amount of the solution(L)

The amount of solute dissolved in grams per liter of the solution is used to determine the solution's strength.
It stands for the solution's potency or concentration.
It uses grams per liter to express.

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A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental

OleMash [197]

Answer:

the empirical (lowest raios) is

C2H4Cl

Explanation:

A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.

What is the empirical formula of this compound?

the compound has ONLY C, H, and Cl

the % Cl = 100% - 24.27% -4.03% = 71.7%

in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H

the number of moles are Cl=71.7/70.91 =1.01= ~ 1

C = 24.27/12.0 = 2.02 =~ 2

H = 403/1.01 = 3.97 =~ 4

so the empirical (lowest raios) is

C2H4Cl

3 0

3 years ago

How many moles of Co2 are produced when 0.2 moles of sodium carbonate reacts with excess HCl

Pie

To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:

2HCl + Na2CO3 = 2NaCl + H2O + CO2

From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:

0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2

Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.

5 0

3 years ago

Which is NOT part of a normal microscope?<br> a.ocular<br> b. nosepiece<br> c.stage<br> d.objective

Evgesh-ka [11]

Answer:

I believe it is a Nose piece

Explanation:

3 0

3 years ago

Read 2 more answers

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),

Semenov [28]

Answer:

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

7 0

3 years ago

Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?

tester [92]

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the <em>empirical formula </em>

EF = K₂O

5 0

3 years ago