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2 years ago

Please help with this!!

Chemistry

1 answer:

loris [4]2 years ago

8 0

Answer:

Limiting: Lab covers. Books produced: 75. Excess amounts: In explanation

Explanation:

Amount that can be produced with just lab covers: 75 books (150/2)

Amount that can be produced with just lined paper: 150 books (7500/50)

Amount that can be produced with graph paper: 120 (3000/25)

Amount that can be produced with staples: Approximately 83 (250/3)

As we can see, the lab covers are limiting as they can only produce 75 books. So, we can only make 75 books.

You will have 3,750 lined paper left over (or 75 books)

(150-75=75 , 75*50=3,750)

You will have 1,125 graph paper left over (or 45 books)

(120-75=45 , 75*25=1,875 , 3000-1875=1125)

And then you will have approximately 25 staples left over (or 8 books)

(83-75=8 , 8*3=225, 250-225=25)

(Hopefully this is correct, I apologize if I messed up)

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The law of conservation of mass says that *

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How many molecular of H2O and O2 are present in 8.5g of H2O2 ?

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2H2O+O2--->2H2O2

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

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azamat

Answer:

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