You removed a Coke can from the refrigerator. When you come back a few minutes later, you realize the can has begun to "sweat".

Assuming constant pressure and temperature, how many moles of gas have been added to the initial 3 moles of gas when the volume

Temka [501]

We first assume that the gas is ideal which is a safe assumption to approximate the answer to the problem. Then we need to know the ideal gas equation and that is:

PV=nRT
where
P- pressure
V- volume
n-number of moles-
R- ideal gas constant
T-temperature.

Since we know that P, T and V are constant, rearranging the equation would lead to:

P/TR = n/V or the ratio of the moles of gas and volume is constant.

(3moles)/2L = (3+x)/4L
where
x is the additional moles.

Solving for x = 3 moles.

8 0

3 years ago

Read 2 more answers

Boyle's Law is about how the volume of a gas is affected by... Boyle's Law is about how the volume of a gas is affected by... te

Lelu [443]

Answer:

please is this a question or answer I don't understand

8 0

2 years ago

How would i calculate molecules of water produced in a reaction?

levacccp [35]

You would calculate them by dividing them and then multiplying to get the final answer

5 0

4 years ago

At catalyst is ______.

AlexFokin [52]

Heya!!!

Answer to your question:

A catalyst is ______
B. Not used up in a reaction.

Catalyst do not change the amount of reactants or products or itself get used but it just change the rate of reaction.

Hope it helps *_*

4 0

4 years ago

Which has the most gravitational potential energy- a 10 kg ball that is 2 meters high or a 10 kg ball that is 4 meters high?

Alenkasestr [34]

Answer:

The 10 kg ball that is 4 meters high is greater than a 10 kg ball that is 2 meters high

Explanation:

From the question,

Gravitational Potential Energy of the first ball

P.E1 = mgh.................... Equation 1

Where P.E1 = Gravitational Potential Energy of the first ball, m = mass of the first ball, h = height of the first ball, g = acceleration due to gravity.

Given: m = 10 kg, h = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E1 = 10×2×9.8

P.E1 = 196 J.

Similarly,

Gravitational potential energy of the second ball is

P.E2 = m'gh'................... Equation 2

Where P.E2 = Gravitational potential Energy of the second ball, m' = mass of the second ball, h' = height of the second ball.

Given; m' = 10 kg, h' = 4 m

Substitute these values into equation 2

P.E2 = 10×4×9.8

P.E2 = 392 J

From the above calculation,

P.E2 is greater than P.E1.

Hence, the 10 kg ball that is 4 meters high is greater than a 10 kg ball that is 2 meters high

3 0

3 years ago