Which term is affected by the shape of a particle?

A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

7 0

4 years ago

PLEASE HELP ME & will mark the brainliest!

AfilCa [17]

Answer:

HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)

Explanation:

HCOOH is a weak acid and so will not ionised completely in solution.

KOH is a strong base and will ionised completely as shown below

KOH(aq) –> K+(aq) + OH-(aq)

The overall reaction can be written as follow:

HCOOH(aq) + K+(aq) + OH-(aq) —> HCOO-(aq) + K+(aq) + H2O(l)

Cancel out the K+ to obtain the net ionic equation as shown below

HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)

3 0

4 years ago

What trend in atomic radius do u see as you go a group family on the periodic table?

exis [7]

Atomic radius increases down a group

7 0

3 years ago

Read 2 more answers

What mass of glucose can be produced from a photosynthesis reaction that occurs using 10 moles of carbon dioxide?

BlackZzzverrR [31]

Hope this help. Answer will be on image.

6 0

4 years ago

Read 2 more answers

If you complete and balance the following oxidation-reduction reaction in basic solution NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(a

andreev551 [17]

Answer:

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

Step 2: Balancing electrons

NO2- → NH3

On the left side N has an oxidation number of +3 and on the right side -3.

NO2- +6e-→ NH3

Al(s) → Al(OH)4-

On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

NO2- +6e-→ NH3

2(Al(s) → Al(OH)4- +6e-)

Step 3: Balance with OH/H2O

NO2- +6e +5H2O → NH3 +7OH-

2Al +8OH- → 2Al(OH)4- + 6e-

Step 4: The netto reaction

NO2- + 5H2O + 2Al + 8OH- → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

6 0

4 years ago