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oksian1 [2.3K]
3 years ago
7

(b) the absolute temperature of the gas at which 3.33x103 mol occupies 478 mL at 750 torr,​

Chemistry
1 answer:
bija089 [108]3 years ago
6 0

Explanation:

T=?

Given

n=3.33×10³

V=478mL=4.78L

P=750torr= 0.987 Atm

simply use the formula

PV =nRT

where R is universal gas constant

0.987×4.78= 3.33×10³×8.314×T

T = 0.000170°c

T= 273.15 K!

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Air is made of nitrogen, carbon dioxide, and small amounts of other gases. What is the pressure of O₂, total is 1 atm if PN₂ = 5
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Answer:

The pressure of O₂ is 0.8 atm.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case:

PT=Pnitrogen + Pcarbon dioxide + Pother gases

Being:

  • Pnitrogen: 593.4 mmHg
  • Pcarbon dioxide: 3 mmHg
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and replacing:

PT= 593.4 mmHg + 3 mmHg + 7.1 mmHg

you get:

PT= 603.5 mmHg

Being 760 mmHg= 1 atm, you get:

PT= 603.5 mmHg= 0.8 atm

<u><em>The pressure of O₂ is 0.8 atm.</em></u>

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3 years ago
Why does the Moon orbit Earth instead of the Sun?
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Gravity depends on distance and the moon is closer to earth
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7.6 The diagrams show the atoms in four different substances. Each circle represents an atom.
Illusion [34]

Answer:

A and C represent elements while B and D represent Compounds

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8 0
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What volume does .0685 mol of gas occupy at STP? 1mole =22.4
GuDViN [60]
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4 0
2 years ago
Read 2 more answers
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
2 years ago
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