Which of the following statements is true about gasses?

If the pressure inside the cylinder increases to 1.3 atm, what is the final

EleoNora [17]

Answer:

1.4 × 10² mL

Explanation:

There is some info missing. I looked at the question online.

The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?

Step 1: Given data

Initial volume (V₁): 215 mL

Initial pressure (P₁): 625 mmHg

Final volume (V₂): ?

Final pressure (P₂): 1.3 atm

Step 2: Convert 625 mmHg to atm

We will use the conversion factor 1 atm = 760 mmHg.

625 mmHg × 1 atm/760 mmHg = 0.822 atm

Step 3: Calculate the final volume of the air

Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL

5 0

3 years ago

Hydrocyanic acid, HCN, is a weak acid. (a) Write the chemical equation for the dissociation of HCN in water. (b) Identify the Br

12345 [234]

Answer: a) $HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)$

b) $HCN$ : acid $CN^-$ :conjugate base.

And, $H_2O$ : base $H_3O^+$ : conjugate acid.

c) $HCN(aq.)+NaOH(aq)\rightleftharoons NaCN(aq.)+H_2O(l)$

d) $NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)$

e) $NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)$

Explanation:

a) Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases $H^+$ ions in their aqueous states.

The equation for the dissociation of $HCN$ acid is given by:

$HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)$

b) According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

$HCN$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $CN^-$ which is a conjugate base.

And, $H_2O$ is gaining a proton, thus it is considered as a base and after gaining a proton, it forms $H_3O^+$ which is a conjugate acid.

c) Neutralization reaction is a reaction in which an acid reacts with base to produce salt and water.

$HCN(aq.)+NaOH(aq)\rightarrow NaCN(aq.)+H_2O(l)$

d) The chemical equation for dissociation of $NaCN$ in water.

$NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)$

e) The chemical equation for the reaction of $NaCN$ and $HCI$

$NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)$

4 0

3 years ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

2 years ago

A gas takes up a volume of 35 L, and has a pressure of 4.8 atm. What is the new pressure of

joja [24]

Answer:

7.3 atm

Explanation:

- Use the formula P1V1 = P2V2

- Rearrange formula and then plug in values.

- Hope this helped! Let me know if you need more help or a further explanation.

3 0

3 years ago

The temperature in your town is 31°F. The radio announcer says that the temperature will drop 15 degrees. What will the temperat

hoa [83]

Answer:

The temperature of the town will be 15°F.

The equation showing the final temperature after drop say x degrees can be written as:

$T'=(T-x)^oF$

Explanation:

The current temperature in our town = T = 31°F

Temperature drop suggested by radio announcer = 15°

Temperature of the town after temperature drop = T'

$T'=T-15^oF$

$T'=31^oF-15^oF=16^oF$

The equation showing the final temperature after drop say x degrees can be written as:

$T'=(T-x)^oF$

3 0

3 years ago