Ask question

Ask question

All categories

Paladinen [302]

4 years ago

15

A thin-walled plastic bottle is sealed at an altitude of 19,000 feet. What would happen to this bottle if it is carried down to

1,000 feet?

2 answers:

Stolb23 [73]4 years ago

8 0

Answer:

It will shrink and collapse.

Explanation:

In the given question it is given that a thin-walled plastic bottle, which is sealed at a height of 19000 feet is carried down to 1000 feet. At an altitude of 19000 feet, the air will be colder, and at the same time, the pressure would be decreasing. Thus, if the bottle is carried down to 1000 feet, the pressure will increase resulting in the collapse and shrinking of the plastic bottle.

Dima020 [189]4 years ago

5 0

The bottle would shrink and collapse

You might be interested in

Problem PageQuestion Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature

Marrrta [24]

Complete question

The complete question is shown on the first uploaded image

Answer:

The value is $K_c = 2.69 *10^{-5}$

Explanation:

From the question we are told that

The equation is

$Fe_20_3_{(s)}+3H_{(g)}\to2Fe_{(s)}+3H_2O_{(g)}$

Generally the equilibrium is mathematically represented as

$K_c = \frac{[H_2O]^2}{[H_2]^3}$

Here $[H_2O]$ is the concentration of water vapor which is mathematically represented as

$[H_2O ] = \frac{n_w}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_w$ is the number of moles of water vapor which is mathematically represented as

$n_w = \frac{m_w}{Z_w}$

Here $m_w$ is the mass of water given as 2.00 g

and $Z_w$ is the molar mass of water with value 18 g/mol

So

$n_w = \frac{2}{18}$

=> $n_w = 0.11 \ mol$

So

$[H_2O ] = \frac{0.11}{8.9 }$

=> $[H_2O ] = 0.01236 \ M$

Also

$[H]$ is the concentration of hydrogen gas which is mathematically represented as

$[H ] = \frac{n_v}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_v$ is the number of moles of hydrogen gas which is mathematically represented as

$n_v = \frac{m_v}{Z_v}$

Here $m_w$ is the mass of water given as 4.77 g

and $Z_v$ is the molar mass of water with value 2 g/mol

So

$n_w = \frac{4.77}{2}$

=> $n_w = 2.385 \ mol$

So

$[H_2O ] = \frac{2.385}{8.9 }$

=> $[H_2O ] = 0.265 \ M$

So

$K_c = \frac{( 0.01236 )^3}{ (0.265 )^2}$

=> $K_c = 2.69 *10^{-5}$

7 0

3 years ago

enot [183]

Could you add a screenshot so we know what's being asked please?

5 0

3 years ago

A gas occupies 49 liters at a pressure of 367 mm Hg. What is the volume when the pressure is increased to 784 mm Hg?

vodka [1.7K]

Answer:

22.94 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have two different values of V and P:

P₁V₁ = P₂V₂

P₁ = 367.0 mm Hg, V₁ = 49.0 L.

P₂ = 784.0 mm Hg, V₂ = ??? L.

∴ V₂ = P₁V₁/P₂ = (367.0 mm Hg)(49.0 L)/(784.0 mm Hg) = 22.94 L.

7 0

3 years ago

Chegg All of the following compounds are soluble in water except ____ . KClO4 Ba(OH)2 KCl PbCl2 AgNO3

EastWind [94]

Answer:

The answer is " $PbCl_2$ "

Explanation:

The water solution of PbCl2 is poor. It also has 2 distinct dissolving properties. In cold water, this will not dissolve easily.It is water-insufficiently soluble. And that in cold water, it is indeed insoluble. That's why we know that all energy should be available for a solid to dissolve in the liquid to counteract that attractive force between the ions in the grid.

6 0

3 years ago

A sample of gas with a mass of 21.3 g is confined to a vessel of volume 7.73 L at 0.880 atm and 30.00C.

rodikova [14]

Answer:

(a) 77.9 g/mol

(b) 3.18 g / L

Explanation:

<u>(a)</u> We need to use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the moles, R is the gas constant, and T is the temperature in Kelvins.

Notice that we don't have moles; we instead have the mass. Remember, though that moles can be written as m/M, where m is the mass and M is the molar mass. So, we can replace n in the equation with m/M, or 21.3/M. The components we now have are:

- P: 0.880 atm

- V: 7.73 Litres

- n: m/M = 21.3 g / M

- R: 0.08206

- T: 30.00°C + 273 = 303 K

Plug these in:

PV = nRT

(0.880)(7.73) = (21.3/M)(0.08206)(303)

Solve for M:

M = 77.9 g/mol

<u>(b)</u> The equation for the molar mass is actually:

M = (dRT)/P, where d is the density

We have all the components except d, so plug them in:

77.9 = (d * 0.08206 * 298) / 1

Solve for d:

d = 3.18 g / L

3 0

3 years ago

Read 2 more answers