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sdas [7]
3 years ago
12

Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of

1.5 atm. 1 L × atm = 101 J.
Chemistry
1 answer:
marta [7]3 years ago
6 0

Answer:

Work done by the system = 4545 J

Explanation:

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,  

\Delta V = 45 - 15 L = 30 L

P = 1.5 atm

w=-1.5\times 30\ atmL

Also, 1 atmL = 101 J

So,  

w=-1.0\times0.75\times 101\ J=-4545\ J (negative sign implies work is done by the system)

<u>Work done by the system = 4545 J</u>

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If 25.0 g NO are produced, how many grams of nitrogen gas are used?
ivanzaharov [21]

Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

       N + O ---------> NO

Mole ratio of N to NO is 1:1

Mole of 25.0 g of NO = 25/30.01 = 0.833 moles

Equivalent mole of N = 0.833 moles

Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

7 0
1 year ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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