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1 year ago

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A 157-mL sample of gas is collected over water at 22°C and 753 torr. What is the volume of the dry gas at STP? (The vapor pressu

re of water at 22°C = 20. torr)

1 answer:

galina1969 [7]1 year ago

3 0

Answer:

Volume at STP = 0.1401 L

Explanation:

We are given:

Vapor pressure of water = 20 torr

Total vapor pressure = 753 torr

Vapor pressure of gas = Total vapor pressure - Vapor pressure of water = 753 torr - 20 torr = 733 torr

Using

$PV=nRT$

where,

P = pressure of the gas = 733 torr

V = Volume of the gas = 157 mL= 0.157 L

T = Temperature of the gas = $22^oC=[22+273]K=295K$

R = Gas constant = $62.3637\text{ L.torr}mol^{-1}K^{-1}$

n = number of moles of gas = ?

Putting values in above equation, we get:

$733 torr\times 0.157L=n\times 62.3637\text{L.torr}mol^{-1}K^{-1}\times 295K\\\\n=\frac{733\times 0.157}{62.3637\times 295}=0.006255mol$

At STP, one mole of the gas occupies a volume of 22.4 L

So, 0.006255 mole of the gas occupies a volume of 22.4*0.006255 L

<u>Volume at STP = 0.1401 L</u>

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Answer:

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Explanation:

Given:

Mass of solid sphere(m) = 6.1 kg

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Volume of the sphere = 6.1 kg/2600 kg/m³ = 0.002346 m³

The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force = $\rho\times gV$

Where

$\rho$ is the density of the fluid

g is the acceleration due to gravity

V is the volume displaced

The free body diagram of the sphere is shown in image.

According to image:

$mg=\rho\times gV+T$

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Applying in the equation to find the density of the liquid as:

$6.1\times 9.81=\rho\times 9.81\times 0.002346+26$

$33.841=\rho\times 9.81\times 0.002346$

$\rho=\frac{33.841}{9.81\times 0.002346}$

$\rho=1470.43 kgm^3$

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2 years ago

Explain how the models you developed show that when methane combines with oxygen to form carbon dioxide and water, no atoms are

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1 year ago

A solution contains 10.0 g pentane, C5H12, 20.0 g hexane, C6H14, and 10.0 g benzene, C6H6. What is the mole fraction of hexane?

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Answer:

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2 years ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

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Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

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Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

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2 years ago

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1 year ago

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