Chemistry, 05.10.2019 09:01 zhellyyyyy

The volume of a gas-filled balloon is 30.0 l at 40 °c and 153 kpa pressure. what volume will the balloon have at standard temperature and pressure (273.15 k and 101.3 kpa)?
a. 17.3 l
b. 23.7 l
c. 39.5 l
d. 51.9 l

a gas that has a volume of 28 liters, a temperature of 45 °c, and an unknown pressure, has its volume increased to 34 liters and its temperature decreased to 35 °c. if i measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?
a. 1.5 atm
b. 1.7 atm
c. 2.8 atm
d. 2.5 atm

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Answer from: alexandria3498

Hello!

The volume of a gas-filled balloon is 30.0 L at 40 °C and 153 kPa pressure. What volume will the balloon have at standard temperature and pressure (273.15 K and 101.3 kPa)?

a. 17.3 L

b. 23.7 L

c. 39.5 L

d. 51.9 L

We have the following data:

V1 (initial volume) = 30 L

T1 (initial temperature) = 40ºC (in Kelvin)

TK = TºC + 273.15

TK = 40 + 273.15 → T1 (initial temperature) = 313.15 K

P1 (initial pressure) = 153 kPa

V2 (final volume) = ? (in L)

T2 (final temperature) = 273.15 K

P2 (final pressure) = 101.3 kPa

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{153*30}{313.15} =\dfrac{101.3*V_2}{273.15}$

$\dfrac{4590}{313.15} =\dfrac{101.3\:V_2}{273.15}$

multiply the means by the extremes

$313.15*101.3\:V_2 = 4590*273.15$

$31722.095\:V_2 = 1253758.5$

$V_2 = \dfrac{1253758.5}{31722.095}$

$\boxed{\boxed{V_2 \approx 39.5\:L}}\:\:\:\:\:\:\bf\blue{\checkmark}$

c. 39.5 L

_______________________

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure, has its volume increased to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

a. 1.5 atm

b. 1.7 atm

c. 2.8 atm

d. 2.5 atm

We have the following data:

V1 (initial volume) = 28 L

T1 (initial temperature) = 45ºC (in Kelvin)

TK = TºC + 273.15

TK = 45 + 273.15 → T1 (initial temperature) = 318.15 K

P1 (initial pressure) = ? (in atm)

V2 (final volume) = 34 L

T2 (final temperature) = 35ºC (in Kelvin)

TK = TºC + 273.15

TK = 35 + 273.15 → T2 (final temperature) = 308.15 K

P2 (final pressure) = 2 atm

Now, we apply the data of the variables above to the General Equation of Gases, let's see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{P_1*28}{318.15} =\dfrac{2*34}{308.15}$

$\dfrac{28\:P_1}{318.15} =\dfrac{68}{308.15}$

multiply the means by the extremes

$28\:P_1*308.15 = 318.15*68$

$8628.2\:P_2 = 21634.2$

$P_2 = \dfrac{21634.2}{8628.2}$

$P_2 = 2.507382768... \to \boxed{\boxed{P_2 \approx 2.5\:atm}}\:\:\:\:\:\:\bf\blue{\checkmark}$

d. 2.5 atm

_______________________

$\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

Answer from: Quest

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.

Answer from: Quest

the answer is c

tell me if wrong

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The volume of a gas-filled balloon is 30.0 l at 40 °c and 153 kpa pressure. what volume will the bal...

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