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katen-ka-za [31]
3 years ago
7

If 14.5 kJ of heat were added to 485 g of liquid water, How much would its temperature increase?

Chemistry
1 answer:
olga55 [171]3 years ago
3 0

<u>Answer:</u> The increase in temperature is 7.14°C

<u>Explanation:</u>

To calculate the change in temperature, we use the equation:

q=mc\Delta T

where,

q = heat absorbed = 14.5 kJ = 14500 J    (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 485 g g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = ?

Putting values in above equation, we get:

14500J=485g\times 4.184J/g.^oC\times \Delta T\\\\\Delta T=7.14^oC

Hence, the increase in temperature is 7.14°C

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Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of O_2 in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}      ........(1)

where,

p_A = partial pressure of oxygen at sea level = ?

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Putting values in equation 1, we get:

p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

p_T=36.5atm\\p_{O_2}=0.209atm

Putting values in equation 1, we get:

0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

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The density of helium is 0.17 g/L. What Is the mass of helium in a 5.4 L helium balloon? (2 sig figs) *
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The correct answer is 0.92 g

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