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4.A 100 L sample of gas is at a pressure of 80 kPa and a temperature of 200 K. What volume does the same

dexar [7]

Answer:

V₂ = 107.84 L

Explanation:

Given data:

Initial volume = 100 L

Initial pressure = 80 KPa (80/101 =0.79 atm)

Initial temperature = 200 K

Final temperature =273 K

Final volume = ?

Final pressure = 1 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁T₂ /T₁P₂

V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm

V₂ =21567 atm.L.K /200 K.atm

V₂ = 107.84 L

8 0

3 years ago

Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls t

Kobotan [32]

Now we know that

Q = mc∆T

Where Q is y energy measured in Joules.

m is the mass measured in grams

c is the specific heat of the substance measured in joule per gram degree Celsius.

∆T is the change in temperature measured in degree Celsius.

Let Q1 be the specific heat of the lighter ball.

c1 be the specific heat of the lighter ball.

m1 be the mass of the lighter ball.

∆T1 be the change in the of the lighter ball.

Let Q2 be the specific heat of the heavier ball.

c2 be the specific heat of the heavier ball.

m2 be the mass of the heavierr ball.

∆T2 be the change in the of the heavier ball.

It has been given that the heat lost, that is Q is the same for both the balls of different mass.Which implies Q1= Q2

Specific heat(c) is the same for both the balls since both are made up of iron. c1=c2

Now heat lost by the lighter ball = heat lost by the heavier ball.

Q1= Q2

m1c1∆T1= m2c2∆T2

Since c1=c2

We get

m1/m2= ∆T2/∆T1

Thus we can say since m2>m1,∆T1> ∆T2.

Now initial temperature of both the balls are 100 degree Celsius.

∆T1 = Final temperature(T1 )-100.

∆T2= Final temperature ( T2)-100

Now since the ∆T1> ∆T2 as arrived from the above equation we can conclude that the final temperature of the ball 1 is greater than that of the ball 2. Since the ball 1 as per our assumption is the lighter ball,the final temperature of the ball which has lighter mass is greater than that of the one having a greater mass.

5 0

3 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

2 years ago

1. a.) Calculate the wavelength of light which has a frequency of 5.25 x 10 14 Hz.

Scilla [17]

<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

= 2.998 × 10^8m/s ÷ 5.25 × 10^14 Hz

= 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

= 5.71 × 10² nm

The wavelength of light wave 5.71 × 10² nm

3 0

2 years ago

On increasing which of the following factors, decreases the extent of physisorption? Surface area of the adsorbent Critical temp

Citrus2011 [14]

Answer: On increasing temperature at which adsorption is carried out decreases the extent of physisorption.

Explanation:

An adsorption where molecules of the adsorbate are placed or held on the surface of adsorbent by Vander waals forces is called physisorption.

There is basically physical bonding between the molecules of gas to the surface of a solid or liquid.

Physisorption is reversible in nature and occurs at low temperatures.

It is not specific in nature which means that all gases are adsorbed on the surface of every solid substance to some extent.

Thus, we can conclude that on increasing temperature at which adsorption is carried out decreases the extent of physisorption.

4 0

2 years ago