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Yuliya22 [10]
3 years ago
7

What is the final pressure (expressed in atm) of a 3.05 L system initially at 724 mm Hg and 298 K, that is compressed to a final

volume of 3.00 L at 273 K?
Chemistry
1 answer:
maksim [4K]3 years ago
5 0

Hey there!:

V1 = 3.05 L

V2 = 3.00 L

P1 = 724 mmHg

P2 = to be calculated

T1 = 298 K

T2 = 273 K

Therefore:

P1*V1  / T1  = P2*V2 / T2

P2 = ( P1*V1 / T1  )   * T2 / V2

P2 = 724 * 3.05 * 273 / 298 * 3.00

P2 = 602838.6 / 894

P2 = 674.31 mmHg

1 atm ----------- 760 mmHg

atm ------------- 674.31 mHg

= 674.31 * 1 / 760

= 0.887 atm

Hope this helps!

You might be interested in
1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
HELP ME NOW ASAP PLEASE, 10 POINTS HURRY
xxMikexx [17]
Which of those depend on mayfly nymphs. In other terms which animal eats mayfly nymphs?

I would say frogs
7 0
3 years ago
Which of the following is most likely to form multiple (double or triple) bonds? 1. Li 2. Cl 3. N 4. H 5. F
alukav5142 [94]

Answer: Option (3) is the correct answer.

Explanation:

Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.

Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.

Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.

Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.

Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.

3 0
3 years ago
3. In a typical titration experiment a student titrates a 5.00 mL sample of formic acid (HCOOH), a monoprotic organic acid, with
ira [324]

Answer:

2.893 x 10⁻³ mol NaOH

[HCOOH] = 0.5786 mol/L

Explanation:

The balanced reaction equation is:

HCOOH + NaOH ⇒ NaHCOO + H₂O

At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.

The moles of base added is calculated as follows:

n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH

Extra significant figures are kept to avoid round-off errors.

Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.

(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH

The concentration of HCOOH to the correct number of significant figures is then calculated as follows:

C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L

The question also asks to calculate the moles of base, so we convert millimoles to moles:

(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH

7 0
3 years ago
An occuled front moves over farmland that has been experiencing drought conditions.What change in weather likely bring?
Gnom [1K]

It is C. Extended period of rain.

Hope this helps! Mark as brainiest please.

8 0
3 years ago
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