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natita [175]
2 years ago
9

An aspirin tablet contains 325.0 mg of aspirin, which has the molecular formula, C9H8O4. How many molecules of aspirin are in th

e tablet?
Chemistry
1 answer:
Inga [223]2 years ago
6 0

Answer:

1.08X10²¹ molecules aspirin are in the tablet

Explanation:

Molar mass of aspirin is 180.15 g/m (C₉H₈O₄)

This data means that 1 mol weighs 180.15 g so:

325 mg = 0.325 g

If 180.15 g are contained in 1 mol of aspirin

0.325 g (which is the weight from the tablet) are contained in ...

(0.325 g  .1m)/180.15 g = 1.80x10⁻³ moles

1 mol has 6.02x10²³ molecules (NA)

1.80x10⁻³ moles are contained by ...

1.80x10⁻³ . NA = 1.08X10²¹ molecules

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When monomers combine to form a condensation polymer, another product is also formed. typically, this product is _____. methanol
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The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t
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The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
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a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
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Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 =<em> 76.5%</em>

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = <em>17.6%</em>
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2 years ago
What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride
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<u>Given:</u>

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

<u>To determine:</u>

The mass of CaCO3 produced

<u>Explanation:</u>

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol  = 75 g

Ans: Mass of CaCO3 produced = 75 g

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2 years ago
Assume the atomic mass of element X is 22.99 amu. A 17.15−g sample of X combines with 14.17 g of another element Y to form a com
Alenkasestr [34]
The formula of the compound is XY. This means that the relation between the moles is 1: 1. One mole of X per one mole of Y.

From the information about the element X you can determine the number of moles of X (which is the same that the number of moles of Y).

# of moles of X = weigth of X / atomic mass of X = 17.15 g / 22.9 g/mol = 0.74598

Now the atomic mass of Y = weight of Y / # of moles of Y = 14.17 g / 0.74598 mol = 19 amu
8 0
2 years ago
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