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3 years ago

if the pressure in a balloon is 20 atm at 20 K, what will be the pressure when the temperature is raised to 30K?

Chemistry

1 answer:

nalin [4]3 years ago

3 0

Answer:

30 atm

Explanation:

P1/P2 = T1/T2

20 atm/P2 = 20K/30K

P2 = 20*30/20 = 30 atm

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3 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

NeTakaya

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 .

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

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

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>

8 0

3 years ago

At standard pressure, ammonia melts at 195 K and boils at 240 K. If a sample of ammonia at standard pressure is cooled from 200

Svetradugi [14.3K]

Answer:

I) the heat capacity of ammonia(s)

II) the heat capacity of ammonia(ℓ)

IV) the enthalpy of fusion of ammonia

Explanation:

Initially, ammonia at 200 K is liquid. To calculate the change of enthalpy from 200 K to 195 K (melting point) we need to know the heat capacity of ammonia(ℓ).

At 195, ammonia is in the transition from liquid to solid (solidification). To calculate the change of enthalpy in that process we need to know the enthalpy of solidification of ammonia, which has the same value but opposite sign to the enthalpy of fusion of ammonia.

From 195 K to 0 K, ammonia is solid. To calculate the change of enthalpy in that process we need to know the heat capacity of ammonia(s).

7 0

3 years ago

This element has the most protons of any element in group 15.?

lozanna [386]

Bismuth... hope it helps!

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3 years ago

How are period and frequency related to each other?

frosja888 [35]

Answer:

D. period is reciprocal of frequency.

Explanation:

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3 years ago

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