Ask question

Ask question

All categories

Alekssandra [29.7K]

2 years ago

11

what would the total pressure of a mixture of fluorine, chlorine, and bromine gases be if the partial pressure are 2.20 atm, 6.7

0 atm, 0.03 atm respectively

1 answer:

astraxan [27]2 years ago

4 0

Answer:

The total pressure would be 8, 93 atm

Explanation:

We apply Dalton's laws, where for a gaseous mixture, the total pressure (Pt) is the sum of the partial pressures (Px) of the gases that make up the mixture.

Pt= Pxa + Pxb+ Pxc....

Pt=2, 20 atm+ 6, 70 atm+ 0,03 atm= 8, 93 atm

You might be interested in

When 5800 joules of energy are applied to a 15.2-kg piece of lead metal, how much does the temperature change by

Firdavs [7]

<span>Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. The heat capacity of aluminum at 25 degrees celsius is 0.9 J/g-C. It is expressed as follows:</span><span>

Heat = mC(T2-T1)
5800 J = 152000(0.90)(</span>ΔT)

ΔT = 0.42 °C change in temperature

8 0

2 years ago

Read 2 more answers

PLEASE ASSIST MEEEEEEE!!!!!!!!! 40 POINTS

lana66690 [7]

Answer:

Wavelength of this beam of light: $\rm 4.39\times 10^{-7}\; m$ .

Explanation:

The speed of light in vacuum is approximately $\rm 2.998\times 10^{8}\;m \cdot s^{-1}$ .

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. $1\rm \; Hz$ means one oscillation per second. The frequency of this particular wave is $\rm 6.83\times 10^{14}\; Hz$ . In other words, there are $6.83\times 10^{14}$ oscillations in each second.

The period of oscillation will be equal to

$\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}$ .

In that period of time, a beam of light in vacuum would have traveled

$\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m$ .

In other words, if this beam of light of frequency $\rm 6.83\times 10^{14}\; Hz$ is in vacuum, its wavelength will be equal to $\rm 4.39\times 10^{-7}\; m$ .

8 0

2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

2 years ago

The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.

kirza4 [7]

Answer:

See the explanation

Explanation:

1) The Lewis structure for $CO_2$ has a central Carbon<em> </em>atom attached to Oxygen atoms.

In the $CO_2$ we will have a structure: O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.

2) These atoms are held together by <u>double bonds.</u>

<u></u>

Again in the structure of $CO_2$ : O=C=O we only have double bonds.

3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.

Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.

4. The carbon atom is <u>Sp</u> hybridized.

We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.

5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.

(See figures)

Figure 1: Carbon hybridization

Figure 2: Oxygen hybridization

6 0

2 years ago