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Rudiy27
3 years ago
7

3. At the end of a warm summer day, imagine you are jumping from the 80°F air into an

Chemistry
1 answer:
djyliett [7]3 years ago
8 0

Answer:

  • <u>1. Since the temperature of your body is higher than the temperature of the air and of the water, heat will flow from your body to the air and pool.</u>

<u></u>

  • <u>2. The pool feels cooler than air because the water can absorb heat energy faster than the air, due to liquids are better conductors than gases.</u>

<u></u>

Explanation:

Heat always flows from warmer substances to colder ones.

The normal body temperature is 98ºF. Therefore, the heat will flow from your body to the air and pool, which are at a lower temperature of 80ºF. In both cases, you will lose thermal energy and the external parts of your body will cool down.

The difference between both cases is in the heat conduction capacity of both air and water.

Liquids (and solids) are better <em>thermal conductors </em>than gases because the conduction of heat occurs as result of the direct contact between the particles of matter: the atoms or molecules in hot matter vibrate faster than their neighbors and transmit them kinetic energy by direct contact.

Therefore, the liquid water in the swimming pool, at the same temperature than the air, will be able to absorb more heat in the same time from the body.

In conclusion, the body will cool down faster in water than in air which is why the pool feels cooler than air.

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Answer:

Option D. pH= 1.3 strong acid

Explanation:

From the question given:

The hydrogen ion concentration [H+] = 0.05 M

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2 years ago
A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
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Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

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\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

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example of oxide containing oxygen :

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