All categories

3 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

1 answer:

7 0

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ 63 g

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

brainly.com/question/4147359?referrer=searchResults

You might be interested in

If there are 1.55 x 1024 molecules of hydrogen peroxide (H2O2), what is the mass of the sample?

bezimeni [28]

Answer:

87.54 g of H₂O₂

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 1.55×10²⁴ molecules

Mass of H₂O₂ =.?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of H₂O₂

Next, we shall determine the mass of 1 mole of H₂O₂. This can be obtained as follow:

1 mole of H₂O₂ = (2×1) + (2×16)

= 2 + 32

= 34 g

Thus,

6.02×10²³ molecules = 34 g of H₂O₂

Finally, we shall determine mass of H₂O₂ that contains 1.55×10²⁴ molecules. This can be obtained as follow:

6.02×10²³ molecules = 34 g of H₂O₂

Therefore,

1.55×10²⁴ molecules

= (1.55×10²⁴ × 34)/6.02×10²³

1.55×10²⁴ molecules = 87.54 g of H₂O₂

Thus, 87.54 g of H₂O₂ contains 1.55×10²⁴ molecules.

8 0

3 years ago

A sample of barium nitrate is placed into a jar containing water. The mass of the barium nitrate sample is 27g. Assume the water

34kurt

So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.

using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water

27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g

therefore,
you need 310.34g of water is in the jar.

5 0

3 years ago

Read 2 more answers

A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres

Zolol [24]

Answer:

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

Explanation:

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Considering an initial state 1 and a final state 2, it is true:

P1* V1= P2*V2

In this case:

P1= 20.1 L
V1= 1520 torr
P2= 760 torr
V2= ?

Replacing:

20.1 L* 1520 torr= 760 torr* V2

Solving:

$V2=\frac{20.1 L* 1520 torr}{760 torr}$

V2= 40.2 L

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

4 0

3 years ago

Researchers want to determine the best temperatures for storing batteries. Describe a possible experiment and list the variables

Andrew [12]

How about putting one battery in the freezer while putting another by a radiator or something that gives off heat. Leave them for an hour, then place them in an object that uses batteries and time how long it takes for it to die: Note: It may take many hours for the battery to fully deplete.

4 0

3 years ago

If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

4 years ago