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podryga [215]
3 years ago
11

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

quirrel scuba diving at a depth of 33 meters, where the pressure is 3.5 atm, how big will the squirrel's balloon be?
Chemistry
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?

Putting values in above equation, we get:

1.0\times 3L=3.5\times V_2\\\\V_2=0.86

Thus the squirrel's balloon will be 0.86 L

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Explanation:

The given data is as follows.

         T_{1} = 100^{o}C,       T_{2} = 10^{o}C

       \Delta H_{vap1} = 2257 kJ/kg,     \Delta H_{vap2} = ?

For water, C_{p} = 4.184 kJ/kg ^{o}C

Formula to calculate heat of vaporization is as follows.

  \Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

Hence, putting the values into the above formula as follows.

\Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

2257 kJ/kg - \Delta H_{vap2} = 4.184 kJ/kg ^{o}C (100 - 10)^{o}C

            \Delta H_{vap2} = 2257 kJ/kg - 376.56 kJ/kg

                                       = 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at 10^{o}C is 1880.44 kJ/kg.

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3 years ago
Give two reasons why a luminous flame is not used for heating purposes. (2mks) a​
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Answer:

The Two reasons are :-

1. Combustion efficiency

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Explanation:

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8 0
3 years ago
In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide
Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

ΣP_g_a_s = P_1+P_2+P_3+...+P_n

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr

Solve for Po2:

P_o_2=(760-582.008-0.285) \ torr = 177.707 \ torr

Thus, the partial pressure of diatomic oxygen is 177.707 torr.

<u><em>If you liked this solution, hit Thanks or give a Rating!</em></u>

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3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
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Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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