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3 years ago

5

A sample of neon gas has a volume of 3.15 liters at a pressure of 0.951 atmospheres. What will be the volume of this sample of t

he pressure were increased to 1.292 atmospheres?

1 answer:

stealth61 [152]3 years ago

7 0

The answer would be 2.32 liters

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3 years ago

What is work and state the law of conservation mass?<br>

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Work is a force causing the movement or displacement of an object

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2. Molecules cannot be created or destroyed in a chemical reaction.

3. Compounds cannot be created or destroyed in a chemical reaction.

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3 years ago

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2 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

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2 years ago

You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat a

Juli2301 [7.4K]

Answer:

$Q = 375\,cal$

Explanation:

The quantity of heat transfered from the jellybean to the water is:

$Q = \rho\cdot V \cdot c\cdot \Delta T$

$Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )$

$Q = 375\,cal$

7 0

2 years ago

Read 2 more answers