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jonny [76]
3 years ago
11

If a sample of oxygen occupies a volume of 85.0 L at a pressure of 67.4 kPa and a temperature of 245K , what volume would the ga

s occupy at 179.6 kPa and 273K?
Chemistry
1 answer:
Firdavs [7]3 years ago
4 0

Answer: V2 = 35.54L

Explanation:

Applying

P1= 67.4, V1= 85, T1= 245, P2= 179.6, V2= ?,. T2=273

P1V1/ T1= P2V2/T2

Substitute and simplify

(67.4*85)/245 = (179.6*V2)/273

V2= 35.54L

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makkiz [27]

The answer is 17.5kg.

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2 years ago
WHAT CAN YOU LEARN OR (LEARNING GOAL) ABOUT MITOSIS AND CELL CYCLE?
Marina86 [1]

Explanation:

At the end of mitosis, the new daughter cells contain the same number of chromosomes as the parent cell. Mitosis enables cellular growth and repair in multicellular organisms.

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3 years ago
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.

6 0
3 years ago
Tell how many atoms of each element are in one molecule of the substance of table salt
Tatiana [17]

Answer:

Explanation:

Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.Aug 15, 2020

5 0
2 years ago
Help me please!!
Scrat [10]
I'm pretty sure it's D. Compounds
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7 0
3 years ago
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