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Lunna [17]
3 years ago
8

Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,00

0. Given that a sample of monodisperse polyacrilonitrile weighs 676.8 g and contains molecules of -(CH2CHCN)n-, calculate n.
Chemistry
1 answer:
pentagon [3]3 years ago
6 0

Answer:

7.68 × 10²⁴

Explanation:

Step 1: Calculate the mass of 1 molecule of the monomer CH₂CHCN

We will get the mass of the monomer by adding the masses of the elements.

mCH₂CHCN = 3 × mC + 3 × mH + 1 × mN

mCH₂CHCN = 3 × 12.01 amu + 3 × 1.01 amu + 1 × 14.01 amu = 53.07 amu

Step 2: Convert the mass of the monomer to grams

We will use the conversion factor 1 amu = 1.66 × 10⁻²⁴ g

53.07 amu × 1.66 × 10⁻²⁴ g/1 amu = 8.81 × 10⁻²³ g

Step 3: Calculate "n"

We will divide the mass of the polymer by the mass of the monomer.

n = 676.8 g / 8.81 × 10⁻²³ g = 7.68 × 10²⁴

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almond37 [142]
<h3><u>Answer</u>;</h3>

A. When a reaction is at chemical equilibrium, a change in the system will cause the system to shift in the direction that will balance the change and help the reaction regain chemical equilibrium.

<h3><u>Explanation</u>;</h3>
  • Le Chatelier's principle states that when a change or a "stress" is placed on a system that is at equilibrium, the system will shift in such a way to relieve that change or stress.
  • The stresses include; changing the concentration of reactants or products, altering the temperature in the system and changing the pressure of the system.
  • Therefore; <u><em>when a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. </em></u>
7 0
2 years ago
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Balancee por tanteo las siguientes ecuaciones químicas. Escriba el nombre a reactantes y productos. H2O5 + H2O ---&gt; HNO3 Na2O
Elan Coil [88]

Answer:

a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)

b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)

Explanation:

Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.

a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.

N₂O₅ + H₂O ⇒ HNO₃

Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.

N₂O₅ + H₂O ⇒ 2 HNO₃

b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.

Na₂O + H₂O ⇒ NaOH

Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.

Na₂O + H₂O ⇒ 2 NaOH

3 0
2 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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BabaBlast [244]

Answer:

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Explanation:

8 0
2 years ago
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brilliants [131]

Answer:

see explanation

Explanation:

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12/1 = 12      15/5 = 3

O₂ is the limiting reactant. P₄ will be in excess when rxn stops.

4 0
2 years ago
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