If one mol of gas = 22.4 at STP how many liters is occupied by 3.65 moles of oxygen gas at STP

1 mole gas at STP = 22.4 L

so volume of oxygen at STP:

1)

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2)

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3)

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4)

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5)

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6)

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7)

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

1)

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2)

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3)

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4)

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5)

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6)

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7)

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

44.8 L of N₂.

Explanation:

We'll begin by writing the balanced equation for the. This is illustrated below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted to produce 2 moles of NH₃.

Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e

Xmol of N₂ = (1 × 4)/2

Xmol of N₂ = 2 moles

Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.

Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:

1 mole of N₂ occupies 22.4 L at STP.

Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L

Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.

44.8 L of N₂.

Explanation:

We'll begin by writing the balanced equation for the. This is illustrated below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted to produce 2 moles of NH₃.

Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e

Xmol of N₂ = (1 × 4)/2

Xmol of N₂ = 2 moles

Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.

Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:

1 mole of N₂ occupies 22.4 L at STP.

Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L

Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.

The correct answer is C. 44.8 L

Explanation:

22.4 × 2 = 44.8

The correct answer is C. 44.8 L

Explanation:

22.4 × 2 = 44.8

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

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, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.