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Oduvanchick [21]
3 years ago
9

What is the volume of 1.9 moles of chlorine gas (Cl2) at 298 K and 1.3 atm?

Chemistry
2 answers:
nataly862011 [7]3 years ago
6 0

Hello!

For this exercise you have to use the <em>Ideal Gas Law Equation, </em> as follow:

PV= nRT

Where:

P= pressure ( in this example, atmosphere (atm) is the pressure unit)

V= volume (which is the value you want to calculate)

n= Mol (as reference to the mass of the gas Cl2)

R = Ideal Gas Constant (for this example is a number 0.0821 atm*L/mol*K)

T= Temperature in Kelvin (K)

So, for find the value of the volume, you just have to replace the values given in the equation as follow:

P= 1.3 atm

V = you don't know, is what you want to know.

n= 1.9 mol

R =0.0821 atmL/molK

T =298 K

Remember, never forget to put the units, because teacher can get down your score if you don't do it. So, back with the values to the equation:

13atm * V = 1.9mol * 0.0821 atmL/molK * 298K

Then, we have to solve the equation to find V value, so we has to send 1.3 atm to divide in the other side of the equation:

V = (1.9mol * 0.0821 atmL/molK * 298K)/ 1.3 atm

V = 35.76 L

And, that's it!

I hope this help you a lot, and good luck.

kap26 [50]3 years ago
3 0

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol

The answer is 43 L if I am correct.

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Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.
Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED']  = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E]   (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED]  (dissociation constant)

C)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633  * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED']                                  E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED']  = 0.0579

5 0
3 years ago
What is the percent composition of the u.s quarter, which has the mass of 5.670g
Vitek1552 [10]
The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100

The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams

Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
6 0
3 years ago
What is the name of this hydrocarbon? 2-butane 2-butene 2-butyne 4-butyne 4-propyne
monitta
2-butane is the correct answer
5 0
2 years ago
Read 2 more answers
8
kirill115 [55]

Answer:

K

Explanation:

The picture that demonstrates a state of matter without a fixed shape or fixed volume is picture K.

Gases are one of the three states of matter without a fixed shape and volume. They simply take up the shape of their containers and are free to move all about.

  • Liquids and solids are known to have fixed shape and volume.
  • Liquids are know for their ability to flow. Figure L
  • Solids are rigid bodies. It is the figure J
7 0
2 years ago
What is the pOH of a 2.1x10^-6 m oh- solution?<br><br> pOH=?
balu736 [363]

Answer:

The answer to your question is: pOH = 5.7

Explanation:

pOH is a measure of the concentration of OH ions.

Data

pOH = ?

Molarity = 2.1 x 10 ⁻⁶ M

Formula

pOH = -log [OH]

Substitution

pOH = -log[2.1 x 10⁻⁶]

pOH = 5.7

and pH = 14 - pOH

      pH = 14 - 5.7

      pH = 8.3

5 0
3 years ago
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