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zlopas [31]
3 years ago
14

A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.

Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

Pressure, P = 67.57 atm

Explanation:

<u>Given the following data;</u>

  • Volume = 0.245 L
  • Number of moles = 0.467 moles
  • Temperature = 159°C
  • Ideal gas constant, R = 0.08206 L·atm/mol·K

<u>Conversion:</u>

We would convert the value of the temperature in Celsius to Kelvin.

T = 273 + °C

T = 273 + 159

T = 432 Kelvin

To find the pressure of the gas, we would use the ideal gas law;

PV = nRT

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

Making P the subject of formula, we have;

P = \frac {nRT}{V}

Substituting into the formula, we have;

P = \frac {0.467*0.08206*432}{0.245}

P = \frac {16.5551}{0.245}

<em>Pressure, P = 67.57 atm</em>

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Answer:

1.788 C DEGREES

Explanation:

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T2 = 274.93 K     = 1.788 C

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Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C
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Answer:

The specie which is oxidized is:- CrO_2^-

The specie which is reduced is:- ClO^-

Explanation:

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X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

For the given chemical reaction:

2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-

The half cell reactions for the above reaction follows:

Oxidation half reaction:  CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-

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8 0
3 years ago
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2 years ago
A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is
vekshin1

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

Q= m\times c\times \Delta T

Q = heat absorbed  = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C      

\Delta T={\text{Change in temperature}}=T_2-T_1

T_1=25^oC

Putting values in above equation, we get:

+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC

Hence, the final temperature of the copper is 95°C.

3 0
3 years ago
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