BRAINLIESTTT ASAP!!! PLEASE HELP ME :)
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<h2>Hello!</h2>
The answer is:
We have that there were produced 0.120 moles of
We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.
The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:
Where,
P is the pressure of the gas.
V is the volume of the gas.
n is the number of moles of the gas.
T is the absolute temperature of the gas (Kelvin).
R is the ideal gas constant (to work with pressure in mmHg), which is equal to:
We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.
We can convert from Celsius degrees to Kelvin using the following formula:
So, we are given the following information:
Now, isolating the number of moles, and substituting the given information, we have:
Hence, we have that there were produced 0.120 moles of
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Explanation:
The given data is as follows.
Volume of water = 0.25
Density of water = 1000
Therefore, mass of water = Density × Volume
=
= 250 kg
Initial Temperature of water () =
Final temperature of water =
Heat of vaporization of water () at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point () in 60 min.
Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg
Heat required to evaporate = Amount of water evapotaed × Heat of vaporization
= 62.5 (kg) × 2133 (kJ/kg)
= kJ
All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec
Therefore, heat supplied per unit time = Heat required/time = = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from to
of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)
= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)
= 125520 kJ
Time required = Heat required / Power rating
=
= 3392 sec
Time required to raise the temperature from to
of 0.25
water is calculated as follows.
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.