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choli [55]
2 years ago
7

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

Chemistry
1 answer:
kramer2 years ago
7 0

Answer:

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy is the measure of the degree of disorderness.

In solids, the entropy is very less compared to liquids and gases.

The entropy order is:

solids<liquids<gases

Among the given substances, water in liquid form has a strong intermolecular H-bond.

So, it has also less entropy.

Next acetic acid.

Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.

HCN has slight H-bonding in IT.

Hence, the entropy order is:

Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)

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How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.
Licemer1 [7]

Answer:

(1) 0.10      (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride  = 0.001 mol x 98.06 g/mol =  0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol  maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0
2 years ago
Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right
svetlana [45]

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

3 0
2 years ago
in a candy factory, the nutty chocolate bars contain 22.0% by mass pecans. If 5.0 kg of pecans were used for candy last Tuesday,
zysi [14]
22.0 is the same as saying that in 100 grams of a chocolate bar, there are 22.0 grams of pecans. or to make it easier because of this problem- 100 Kilograms of a chocolate bar, there is 22.0 Kg of pecans. we can use this as a conversion factor (what is used to convert a value to another value. 

conversion factor---> 22.0 kg of pecan= 100 kg of chocolate bar
Note: remember this, what you are converting from goes in the denominator, what you converting to goes in the numerator. 

5.0 Kg of pecan (100 Kg of chocolate bar/ 22.0 Kg of pecan)= 23 Kg of chocolate bar


7 0
2 years ago
A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav
dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of _{19}^{40}\textrm{K} radioisotope follows:

Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles  

Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles  

_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e

By the stoichiometry of above reaction,

1 mole of _{40}^{91}\textrm{Zr} is produced by 1 mole _{41}^{91}\textrm{Nb}

So, 0.17 moles of _{40}^{91}\textrm{Zr} will be produced by = \frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}

Amount of _{82}^{212}\textrm{K}

decomposed will be = 0.17 moles

Initial amount of _{40}^{91}\textrm{Nb}  will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = 0.024

a_o = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

0.024=\frac{0.194}{2^n}

n=3

Therefore, 3 half lives have passed.

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2 years ago
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