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

Respuesta :

Ankit

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!

samueladesida43

The absolute temperature of the gas at which 3.33x10³ mol occupies 478 mL at 750 torr is 1.726 × 10-³K

HOW TO CALCULATE TEMPERATURE:

  • The temperature of a substance can be calculated using the ideal gas law equation as follows:

PV = nRT

Where;

  1. P = pressure (atm)
  2. V = volume (L)
  3. n = number of moles (mol)
  4. R = gas law constant (0.0821 molK/Latm)
  5. T = temperature (K)

According to this question,

  1. P = 750torr = 0.987 atm
  2. V = 478mL = 0.478L

  • 0.987 × 0.478 = 3.33 × 10³ × 0.0821 × T

  • 0.472 = 273.39T

  • T = 0.472 ÷ 273.39

  • T = 1.726 × 10-³K

  • Therefore, the absolute temperature of the gas at which 3.33x10³ mol occupies 478 mL at 750 torr is 1.726 × 10-³K

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