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RSB [31]
3 years ago
9

A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 11.0 mL , how many mi

lliliters of active ingredient are in the bottle?
Chemistry
2 answers:
Alexandra [31]3 years ago
8 0
2.8 HOPE I HELPED!! Sry Caps.
Marta_Voda [28]3 years ago
4 0
All you need to do is change 5% into a decimal which would be 5/100 = .05
then multiply the decimal by the total volume to get the amount of active ingredients in ml
.05 * 56ml = 2.8 ml of active ingredient.

Hope that helps!
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Which question must be answered to complete the table below?
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I need more details so I can answer it for you
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4 years ago
Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0
3 years ago
5. Given the following equation: 8 Fe S8 8 Fes, what mass of iron is needed to react with16.0 grams of sulfur? How many grams of
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18 grams of fes is produced

5 0
4 years ago
URGENT HELP
sdas [7]
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7 0
3 years ago
A 7.300 gram sample of aluminum was quantitatively combined with a sample of selenium to form a compound. The compound weighed 3
adelina 88 [10]

Answer:

The empirical formula is Al2Se3  

Explanation:

<u>Step 1:</u> Data given

Mass of the sample of aluminium = 7.300 grams

Mass of the compound = 39.35 grams

Molar mass of aluminium = 26.98 g/mol

Molar mass of selenium = 78.96 g/mol

<u>Step 2:</u> Calculate mass of selenium

Mass of selenium = mass of compound - mass of aluminium

Mass of selenium = 39.35 - 7.3 = 32.05 grams

<u> Step 3:</u> Calculate moles of Al

Moles Al = Mass Al/ Molar mass Al

Moles Al = 7.300 grams / 26.98 g/mol

Moles Al = 0.2706 moles

<u>Step 4:</u> Calculate moles of Se

Moles Se = Mass Se / Molar mass Se

Moles Se =32.05 g / 78.96 g/mol

Moles Se = 0.4059 moles

<u>Step 5</u>: Divide through the smallest amount of mol

Aluminium: 0.2706 / 0.2706 = 1

Selenium: 0.4059/0.2706 = 1.5

This means for each mol aluminium, we have 1.5 moles of selenium

For each 2 moles of aluminium, we have 3 moles of selenium

The empirical formula is Al2Se3  

This is aluminium(III) selenide.

8 0
3 years ago
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