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3 years ago

Explain how a redox reacation involves electrons in the same way that a neutralization reaction involves protons

1 answer:

8 0

For the neutralization process: an acid acts as a donor and donates protons to the base. On the other hand, the base acts as an acceptor and accepts the transferred protons. In a nutshell, neutralization is mainly proton transfer process.

As for the redox process: the oxidized material usually transfers electrons to the reduced material. In a nutshell, redox is mainly electron transfer process.

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Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K Ho

Maru [420]

Answer : The mass of $CS_2$ is, 555.028 grams

Explanation :

First er have to calculate the concentration of $S_2$ .

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L$

Now we have to calculate the concentration of $CS_2$ .

The given balanced chemical reaction is,

$S_2(g)+C(s)\rightleftharpoons CS_2(g)$

Initial conc. 1.51 0 0

At eqm. conc. (1.51-x) (x) (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CS_2]}{[S_2]}$

Now put all the given values in this expression, we get :

$9.40=\frac{x}{(1.51-x)}$

By solving the term 'x', we get :

x = 1.365 M

Concentration of $CS_2$ = x M = 1.365 M

Now we have to calculate the moles of $CS_2$ .

$\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole$

Now we have to calculate the mass of $CS_2$ .

Molar mass of $CS_2$ = 76 g/mole

$\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g$

Therefore, the mass of $CS_2$ is, 555.028 grams

3 0

4 years ago

What kind of ions do a sodium atom and a chlorine atom become when a valence electron is transferred from one to the other?

kolezko [41]

Answer:

The sodium atom becomes a positive ion and the chlorine atom becomes neg.

Explanation:

Think about this answer

5 0

3 years ago

Read 2 more answers

What is the molarity of the 8.76 g of Na2S dissolved in 0.500 L of solution?

AlexFokin [52]

Answer:

.224 M

Explanation:

To begin, we need to find the molar mass of Na2S

Na=22.99g, S=78.05g

2(22.99)+1(32.07)=78.05

To find Molarity, we use the equation M=moles/Liters of solution

We are given grams, so we must divide by the molar mass of Na2S to find moles.

8.76g Na2S / 78.05g/mol Na2S= .112 moles

Now we use our Molarity equation:

.112 moles Na2S / .500L of solution = .224 M

6 0

2 years ago

Two people are trying to push a large box across a floor. Each person pushes with an equal amount of force. The total amount of

Charra [1.4K]

Answer:

C- 250+250= 500 newtons (the amount the both exerted) :)

Explanation:

5 0

3 years ago

Read 2 more answers

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate

katen-ka-za [31]

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ $H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$ reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$

4 0

3 years ago