CHEMISTRY COLLEGE

What is the difference between a volume that is delivered and a volume that is contained?

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Answer 1
Answer:

Answer:

When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

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Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answers

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released  by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50  x 4.18 x ( 22 - 16.5 )  ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )  

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH =  18.392 kJ / mol

The reaction of ethyl acetate with sodium hydroxide, CH3COOC2H5(aq)+NaOH(aq)⇌CH3COONa(aq)+C2H5OH(aq) is first order in CH3COOC2H5 and first order in NaOH. If the concentration of CH3COOC2H5 was increased by half and the concentration of NaOH was quadrupled, by what factor would the reaction rate increase?

Answers

Answer:

so the reaction rate increases by a factor 6.

Explanation:

For the given equation the reaction is first order with respect to both ester and sodium hydroxide

So we can say that the rate law is

Rate(initial)=K[NaOH][CH_(3)COOC_(2)H_(5)]

now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration

The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.

The new rate law is

Rate(final)=K[1.5XNaOH][4XCH_(3)COOC_(2)H_(5)]

the final rate = 6 X initial rate

so the reaction rate increases by a factor 6.

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

Answer:

The activation energy is =8.1\,kcal\,mol^(-1)

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_(A)=kC_(A)

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^(3)

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_(o)=5dm^(3)s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= (v_(0))/(k)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]

Rearrange the formula is as follows.

k= (v_(0))/(V)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_(o)} is 1.

\epsilon =y_{A_(o)}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^(-1)

The rate constant in case of the CSTR can be calculated by using the formula.

(V)/(v_(0))= (X(1+\epsilon X))/(k(1-X)).............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_(o)} is 0.5

\epsilon =y_{A_(o)}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)

Therefore, the rate constant in case of CSTR comes out to be 2.8s^(-1)

The activation energy of the reaction can be calculated by using formula

k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))

Substitute the all values.

=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)

=8.1\,kcal\,mol^(-1)

Therefore, the activation energy is =8.1\,kcal\,mol^(-1)

How can you remove sand from salt? Which physical property of sand was used in the process?

Answers

Answer:

You could collect the mixture and pour it in water, stir it , ad filter out the sand. This uses the physical property of solubility.

Explanation:

The salt dissolved, the sand didn't.

The distance from Earth to the Moon is approximately 240.000 mi Part C The speed of light is 3.00 x 10 m/s How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.

Answers

Answer:

2.6 sec

Explanation:

The distance between the Earth and the moon = 240,000 miles

Also,

1 mile = 1609.34 m

So,

Distance between the Earth and the moon = 240,000 ×  1609.34 m = 386241600 m

Speed of the light = 3 × 10⁸ m/s

Distance = Speed × Time.

So,

Time = Distance / Speed = 386241600 m / 3 × 10⁸ m/s = 1.3 sec

For back journey = 1.3 sec

So, total time = 2.6 sec

How many joules are released when one atom of americium 241, the isotope used in ionization-type smoke detectors, undergoes apha emission?

Answers

When Americium (Am-241) undergoes alpha decay(He-4) it forms neptunium (Np-237) based on the following pathway:

²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂

The energy released in given as:

ΔE = Δmc²

where Δm = mass of products - mass of reactants

                 = [m(Np-237) + m(He-4)] - [m(Am-241)]

                 = 237.0482+4.0015-241.0568 = -0.0073 g/mol = -7.3 * 10⁻⁶ kg/mol

ΔE = -7.3*10⁻⁶ kg/mol * (3*10⁸ m/s)² = -5.84*10¹¹ J/mol
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