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Temka [501]
3 years ago
10

If 3.11 mol of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 L, what is the temperature of the sample in degrees

Celsius?
Chemistry
1 answer:
nadezda [96]3 years ago
7 0

Answer: 617.35 °C

Explanation:

Using the ideal gas law,

PV = nRT

P = Pressure in (atm or pascal)

V = Volume in (litres or cubic metres)

n = number of moles.

R = Gas constant in (0.0821L .atm/mol.K or 8.314J/mol/K)

T = Temperature in degree Kelvin or Celsius.

PV = nRT

P = 2.91atm, V= 78.13L , n= 3.11

R = 0.0821L .atm/mol.K

T = ?

T = 2.91 x 78.13 / 3.11 x 0.0821 = 227.3583/ 0.255331 = 890.5 K

Converting from Kelvin to Celsius, using: °C = K - 273.15

= 890.5 - 273.15

°C = 617.35

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The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL

<h3>Balanced equation </h3>

CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O

From the balanced equation above,

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<h3>How to determine the volume of NaOH</h3>
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MaVa / MbVb = nA / nB

(0.204 × 46.79) / (0.242 × Vb) = 1

Cross multiply

0.242 × Vb = 0.204 × 46.79

Divide both side by 0.242

Vb = (0.204 × 46.79) / 0.242

Vb = 39.44 mL

Thus, the volume of NaOH needed for the reaction is 39.44 mL

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brainly.com/question/14356286

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The given data is as follows.

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Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

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As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

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