Why do scientists use stirbars in the laboratory?a. to cool a solution
b. to heat a solution
c. to decrease the surface area of a solute
d. to agitate a solution

Answer:

The Carnot engine operates based on the principles of the Carnot cycle, which is a theoretical idealized thermodynamic cycle. To calculate the work done by the engine, we need to use the formula for the efficiency of the Carnot cycle.

The efficiency of a Carnot engine is given by the equation:

Efficiency = 1 - (T2 / T1),

where T2 is the exhaust temperature in Kelvin and T1 is the burn temperature in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin.

The burn temperature is 1957 ˚C, so we add 273 to convert it to Kelvin:

T1 = 1957 + 273 = 2230 K.

The exhaust temperature is 500 ˚C, so we add 273 to convert it to Kelvin:

T2 = 500 + 273 = 773 K.

Now we can calculate the efficiency:

Efficiency = 1 - (T2 / T1) = 1 - (773 / 2230).

Next, we need to calculate the heat input, which is the energy released by burning 1 kg of methane.

The energy released by burning methane can be calculated using the heat of combustion of methane, which is -891 kJ/mol.

To convert this to joules per kilogram, we need to know the molar mass of methane, which is 16 g/mol.

1 kg of methane is equal to 1000 g, so the number of moles of methane in 1 kg is:

1000 g / 16 g/mol = 62.5 mol.

The heat released by burning 1 kg of methane is:

-891 kJ/mol * 62.5 mol = -55,687.5 kJ.

To convert this to joules, we multiply by 1000:

-55,687.5 kJ * 1000 = -55,687,500 J.

Now we can calculate the work done by the engine:

Work = Efficiency * Heat input.

Substituting the values we calculated:

Work = (1 - (773 / 2230)) * (-55,687,500 J).

Finally, we can calculate the work done by the engine in joules.

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Explanation:

First, calculate the moles of using ideal gas equation as follows.

                PV = nRT

or,          n =

                =      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density =

Hence, mass of water will be as follows.

                Density =

             0.998 g/ml =    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles =

                              =              

                              = 0.180 mol

Moles of hydrogen = = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles =

          0.183 mol =              

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles =

                               =              

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is .

Thus, we can conclude that empirical formula of the given compound is .            

Answer is: (1) equal to the total number of electrons gained.

In oxido-reduction reaction, at least one element lose and one element gain electrons.

Oxidation reaction is increasing of oxidation number of atom, because element lost electrons in chemical reaction.

Reduction is lowering oxidation number because atom gain electrons.

For example, in oxidation-reduction reaction: 2H₂ + O₂ → 2H₂O; hydrogen is oxidized (change oxidation number from 0 to +1) and oxygen i s reduced (change oxidation number from 0 to -2).

Hydrogen lost .four electrons anf oxygen gain four electrons

Answer:

B

Explanation:

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