Why do we have rain........................

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

4 years ago

Balance it by oxidation number method:Zn +HNO3----Zn(NO3)2+NO+H2O

timofeeve [1]

Answer:

Answer:

step 1:balance skeleton equation the chemical equation:

Zn +HNO3➔Zn(NO3)2+NO+H2O

step 2: identity undergoing oxidation or reduction

here

Zn➔Zn(NO3)2

Zn is oxidized from 0 to 2 in oxidation no.

HNO3➔NO

N is reduced from 5 to 2 in oxidation no

Step 3: calculate change in oxidation no.

change in oxidation no

in Zn=0-2=-2=2

in

N=5-2=3

Step 4: Balance it by doing crisscrossed multiplication

we get;

3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O

step 6:Balance other atoms except H & O

3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O

3Zn +2HNO3+6HNO3➔3Zn(NO3)2+2NO+H2O

finally: balance H

3Zn +8HNO3➔3Zn(NO3)2+2NO+4H2O

3 0

3 years ago

Read 2 more answers

Which of the following is accurate in describing the placement and classification of iodine?

fenix001 [56]

You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.

4 0

3 years ago

What property of metals is reactivity

nata0808 [166]

The reactivity of a metal is determined by these things.
Firstly, the number of electrons in the outer shell; the fewer the number of electrons in the outer shell, the more reactive the metal.
The number of electron shells also affects reactivity, the more electron shells there are, the more reactive the metal.

7 0

3 years ago

Read 2 more answers

How many moles of Fluorine (F2) are needed to completely react 8.0 moles of NF3?

Anna [14]

Answer:

12 moles of F₂

Explanation:

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

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e

Xmol of F₂ = (3 × 8)/2

Xmol of F₂ = 12 moles

Thus, 12 moles of F₂ is needed for the reaction.

8 0

3 years ago