CHEMISTRY COLLEGE

What is the polarity of a Na-Br bond?nonpolar covalent

definitely polar covalent

likely ionic

slightly polar covalent

slightly ionic

Answers

Answer 1
Answer:

Answer:

likely ionic

Explanation:

The bond between , Sodium and Bromine , in sodium bromide salt , is ionic in nature .

Since ,

The difference in the electronegativity of the two atom is high .

According to Pauling scale ,

The  Electronegativity of Bromine = 2.96

The  Electronegativity of sodium = 0.93

Hence ,

The Difference in electronegativity = ( 2.96 - 0.93 ) = 2.03

Hence ,

The electronegativity difference value is more than 1.8 ,

Therefore,

The bond is ionic in nature .

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The theoretical yield for CuCO3.Cu(OH)2+2h2SO4 ®2CuSO4+2H2O+3CO2

Answers

Answer:

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An alloy is a mixture of metals to obtain a more durable and strongersubstance.
True
False

Answers

Answer: False

Explanation: Alloys are harder and stronger because the different-sized atoms of the mixed metals make the atomic layers less regular, so they cannot slide as easily.

Benzene is a starting material in the synthesis of nylon fibers and polystyrene (styrofoam). Its specific heat capacity is 1.74 J/g·°C. If 16.7 kJ of energy is absorbed by a 225-g sample of benzene at 20.0°C, what is its final temperature?

Answers

Answer: The final temperature of the sample is 62.66°C

Explanation:

To calculate the amount of heat absorbed, we use the equation:

Q=mc\Delta T

where,

Q = heat absorbed = 16.7 kJ = 16700 J   (Conversion factor:  1 kJ = 1000 J)

m = Mass of the sample = 225 g

c = specific heat capacity of sample = 1.74J/g.^oC

\Delta T = change in temperature = T_2-T_1=(T_2-20.0)

Putting values in above equation, we get:

16700=225g* 1.74J/g.^oC* (T_2-20)^oC\n\nT_2=62.66^oC

Hence, the final temperature of the sample is 62.66°C

For the following reaction at equilibrium, which gives a change that will shift the position of equilibrium to favor formation of more products? 2NOBr(g) 2NO(g) + Br 2(g), ΔHº rxn = 30 kJ/mol

Answers

Answer:

Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.  

Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.  

Which of the following is true for the quantum mechanical atomic model? A. Atoms absorb or emit electrons from the nucleus when they interact with electromagnetic radiation.

B. Every atom absorbs all wavelengths of light energy or electromagnetic radiation.

C. Electrons give off electromagnetic radiation when they jump from a high to a low energy level.

D. Electrons are perfectly evenly distributed throughout the atom.​​

Answers

Answer: C. Electrons give off electromagnetic radiation when they jump from a high to a low energy level.

Explanation:

 Electrons give off electromagnetic radiation when they jump from a high to a low energy level in the quantum mechanical atomic model. This is known as the emission spectrum of an atom, and each element has its unique emission spectrum. This phenomenon was explained by the Bohr model of the atom and is a fundamental concept of the quantum mechanical atomic model.

 Option A is incorrect because atoms cannot absorb or emit electrons from the nucleus when they interact with electromagnetic radiation. Option B is also incorrect because atoms only absorb certain wavelengths of light energy or electromagnetic radiation, which corresponds to the energy difference between electron energy levels. Option D is incorrect because electrons are not evenly distributed throughout the atom in the quantum mechanical atomic model; instead, they occupy specific energy levels or orbitals.

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)
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