Answer:
The molecules of a substance which must have the ability to move past one another are said to be flexible.

Explanation:
Those substances are said to be flexible which can be bent without breaking. There are many substances which are hard in nature but still can be bent. The hardness of such materials is due to strong interactions between the molecules and the flexibility comes due to their amorphous backbone. Therefore, greater the crystalline level of macromolecules lesser is the flexibility and greater the amorphous character greater is the flexibility and vice versa. Also, the flexibility of polymers is increased by adding plastisizers in it. Plastisizers make the hard polymers flexible by breaking the crosslinkers and enabling the macromolecules to move past one another.

6 0

3 years ago

If you try to balance an equation by changing subscripts you change...

aksik [14]

Answer:

kwkrofofoxosowoqoaododpdprofpcoxozoskawkdjdn

Explanation:

sklwlrlfclxoskkekrdododosoekekrkrododowoekekfkdodkwkeororkdkdkwejrjrkfidiwi3jr

8 0

2 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

3 years ago

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal

rosijanka [135]

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-511.3kJ/mol=-511300.0J/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of $\Delta n_gRT$ = 0

R = gas constant = 8.314 J/mol.K

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above formula, we get

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K$

$\Delta U=-511300.0J/mol+4955.144J/mol$

$\Delta U=-506344.856J/mol=-506.3kJ/mol$

Therefore, the internal energy change is -506.3 kJ/mol

6 0

3 years ago