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kupik [55]
2 years ago
9

Each molecule of lycopene contains 40 atoms of carbon (plus other atoms). The mass percent of carbon in lycopene is 84.49%. What

is the molar mass of lycopene?a. 1137.2 g/mol
b. 405.9 g/mol
c. 480.4 g/mol
d. 473.4 g/mol
e. 568.6 g/mol
Chemistry
1 answer:
IrinaK [193]2 years ago
5 0

Answer: 568g/mol

Explanation:

It should be noted that there are 40 atoms of carbon in lycopene.

Since mass of 1 carbon = 12g/mol

Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol

Let the molar mass of lycopene be represented by x.

Therefore the molar mass of carbon = x × mass percent of carbon in lycopene

x × 84.49% = 480g/mol

x × 0.8449 = 480g/mol

x = 480/0.8449

x = 568g/mol

The molar mass of lycopene is 568g/mol

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The neutralization reaction that you are studying is the acid-base reaction:

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Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:

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Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
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Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

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