The movement of water across a membrane is termed

I need help with #2. I don't know what law to use for problem.

bagirrra123 [75]

This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.

4 0

3 years ago

PLEASE PLEASE HELP ME!! I WILL BRAINLIST YOU!! A 28.0 g sample of N2 is in a rigid 4.50 L container at 32 °C. Calculate the pres

chubhunter [2.5K]

Answer:

P=4184.36 torr

Explanation:

For this problem we can use the idea gas law,

$PV=nRT$

where P is pressure, V is volume, n is moles of substance, R is the constant, and T is temperature, We will need to manipulate it and a couple values so that we can get our answer in torr. To do this, let's rearrange the equation to solve for P.

$P=\frac{nRT}{V}$

Next, let's convert T (32) to kelvin. To do this, add 273 to the value.

$32+273=305$

Next, let's convert 28g of N2 to moles of N2. To do this, divide 28 by the molar mass of N2 (28.02)

$\frac{28.0}{28.01} \\=.99$ moles of N2

Next, we need to find out what value of R we need to use. Because you want your answer in torr, we will need to use the value of

$R=62.36\frac{torr}{mol*K}$

Now that we have all of our values, let's plug them into the equation (I'll be excluding units for simplicity but they cancel out to leave units of torr in the answer)

$P=\frac{(.99)(62.36)(305)}{4.50}\\P=4184.36$

P=4184.36 torr

<em>Very briefly</em>, I don't know what your periodic table looks like, but mine has nitrogen's molar mass as 14.01. If you have a different mass of nitrogen, this pretty drastically impacts the value. For example, if your nitrogen's molar mass was rounded to 14.0 (making N2 28g) then it would be one mole instead of .99 moles, raising the answer to 4228.70 torr. If you have any different values just plug them in in their proper spots. A similar concept applies to the Kelvin conversion.

6 0

3 years ago

Why is carbon ideal for making different compounds?

In-s [12.5K]

Carbon is the only element that can form so many different compounds because each carbon atom can form four chemical bonds to other atoms, and because the carbon atom is just the right, small size to fit in comfortably as parts of very large molecules.

3 0

3 years ago

Read 2 more answers

1kg of hydrogen is burned in oxygen how much energy is released

Simora [160]

Answer:

121 million joules energy is released

Explain :

As there are 500 moles of hydrogen gas in a kilogram, this means that burning a kilogram of hydrogen gas releases 500 times as much energy or 121 million joules .

6 0

2 years ago

An industrial process of producing ammonia gas runs with a 74.0% yield. How many kilograms of ammonia gas will be produced in th

Tcecarenko [31]

Answer:

37.1 kilograms of ammonia gas will be produced in this process

Explanation:

The percentage yield of the reaction is given by:

$(Yield)\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

According to question

The percentage yield of the given industrial process = 74.0%

The given theoretical yield of ammonia gas = 50.1 kg

The experimental yield of ammonia gas = x

The percentage yield of the reaction is calculated a:

$74.0\%=\frac{x}{50.1 kg}\times 100$

Solving for x, we get:

$x = 37.074 kg\approx 37.1 kg$

37.1 kilograms of ammonia gas will be produced in this process

4 0

3 years ago