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mihalych1998 [28]

2 years ago

15

1. A sample of polystyrene is found to have a number-average molar mass of 89,440 g mol−1 . Neglecting contributions from end gr

oups, calculate the number-average degree of polymerization of this sample. Assuming that the sample has a molar mass dispersity of 1.5, calculate its weight-average molar mass.

1 answer:

Snowcat [4.5K]2 years ago

8 0

Answer:

The weight-average molar mass of polystyrene is 134,160 g/mol.

Explanation:

Molar mass of the monomer styrene , $C_8H_8$ , M=104 g/mol

Given , number average molar mass of the polymer , M'= 89,440 g/mol

Degree of polymerization = n

$n=\frac{M'}{M}=\frac{89,440 g/mol}{104 g/mol}=860$

The weight-average molar mass = $M_{avg}=?$

Molar mass dispersity is ratio of weight-average molar mass to the number average molar mass of the polymer.

$\text{Molar mass dispersity}=\frac{M_{avg}}{M'}$

$1.5=\frac{M_{avg}}{89,440 g/mol}$

$M_{avg}=89,440 g/mol\times 1.5 = 134,160 g/mol$

The weight-average molar mass of polystyrene is 134,160 g/mol.

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Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

2 years ago

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

3 0

2 years ago

Zachary adds 9.7 g to 1.114 g.<br> How many significant figures should his answer have?

makvit [3.9K]

Always use least amount of sig figs possible. So this 9.7 would be (answer): 2 sig figs

5 0

2 years ago

A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of

mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.125

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.125=e^{-0.00440\times t}$

t = 473 year

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