Standard temperature and pressure (stp) are considered to be 273 k and 1.0 atm. predict which of the following changes will cause the volume of the balloon to increase or decrease assuming that the temperature and the gas filling the balloon remain unchanged.

drag the appropriate items to their respective bins.

items:

balloon filled with helium under water at 1.15 atm is released and floats to the surface, which is at stp.

balloon filled with helium at stp floats into the atmosphere where the pressure is 0.5 atm.

balloon filled with helium at stp is submerged under water where the pressure is 1.25 atm.

balloon filled with helium at stp floats into air where the pressure equals 1 atm.

bins:

volume increases

volume decreases

volume is unchanged

part c

a certain gas is present in a 15.0 l cylinder at 3.0 atm pressure. if the pressure is increased to 6.0 atm the volume of the gas decreases to 7.5 l . find the two constants ki, the initial value of k, and kf, the final value of k, to verify whether the gas obeys boyle’s law by entering the numerical values for ki and kf in the space provided.

express your answers to two significant figures separated by a comma.

part d

if a certain gas occupies a volume of 14 l when the applied pressure is 7.0 atm , find the pressure when the gas occupies a volume of 3.5 l .

express your answer to two significant figures and include the appropriate units.

Part A) case 1: volume increases, case 2: volume increases , case 3: volume decreases, case 4: volume is unchanged

Part B) Ki = 45 lt*atm , Kf = 45 Lt*atm

Part C) P₂ = 28 atm

Explanation:

Part A) For this case, we can consider Boyle-Mariotte Law, which links volume and pressure for a gas, this is: with constant temperature and mass for a certaing gas, volume and pressure are inversely proportional : P ∝ 1 / V or, expresing the same: PV = k (k is a constant)

With this said, we may say for 2 different conditions (with temperature and mass constants): P₁V₁ = P₂V₂ = k

For the first case, we can use this formula to predict the final volume of the ballon: V₂/V₁ = P₁/P₂

Case 1 : ballon filled with Helium under water at 1.15 atm is released and floats to the surface, which is STP

P₁ = 1.15 atm , P₂ = 1 atm (STP) ⇒ V₂/V₁ = 1.15/1=1.5 so V₂ > V₁ volume increases

Case 2 : Ballon filled with Helium at STP  floats into the atmosphere where the pressure is 0.5 atm

P₁ = 1 atm (STP), P₂ = 0.5 atm ⇒ V₂/V₁ = 1/0.5 = 2 so V₂ =2V₁ (final volume is twice in volume than original volume) volume increases

Case 3 : Ballon filled with Helium at STP is submerged under water where the pressure is 1.25 atm

P₁ = 1 atm (STP), P₂ = 1.25 atm ⇒ V₂/V₁ = 1/1.25 =0.8 so V₂ < V₁ volume decreases

Case 4 : Ballon filled with Helium at STP floats into air where the pressure equals 1 atm

P₁ = 1 atm (STP) , P₂ = 1 atm ⇒ V₂/V₁ = 1/1 = 1 V₁ = V₂ volume remains constant (is unchanged)

Part B) For a certain gas: V₁  = 15.0 Lt , P₁ = 3.0 atm ; V₂ = 7.5 Lt , P₂ = 6.0 atm

Then we can verify Boyle-Mariotte´s Law, this is, at mass and temperature being constant, P and V are inversely proportional:

PV = k

P₁V₁ = (15.0Lt)*(3.0 atm) = 45 Lt*atm = k i

P₂V₂ = (6.0 atm)*(7.5 atm) = 45 Lt*atm = k f

Part C) For a certain gas, V₁ = 14 Lt, P₁ = 7.0 atm , V₂ = 3.5 Lt , P₂ = ?

Using Boyle-Mariotte´s Law :

P₁V₁ = P₂V₂ ⇒ P₂ = (P₁V₁)/V₂ = (7.0 atm* 14 Lt)/3.5 Lt ⇒ P₂ = 28 atm

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

"the periodic table is arranged in order of atomic number."