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3 years ago

Balance the following equation by redox method H2S + H2 O2 gives S + H2O

1 answer:

5 0

$\text{H}_2\stackrel{-2}{\text{S}} + \text{H}_2\stackrel{-1}{\text{O}_2} \to \; \stackrel{0}{\text{S}} + 2\; \text{H}_2\stackrel{-2}{\text{O}}$

<h3>Explanation</h3>

Find the oxidation state for each element.

The oxidation state of H is +1 in most compounds.

The S atom in H₂S is bonded to two H atoms. The oxidation state of S in H₂S will be -(2 × (+1)) = -2.
Two O atom in H₂O₂ are bonded to two H atoms. On average, the oxidation state for each O atom in H₂O₂ will be - 1/2 × (2 × (+1)) = -1.
S has an oxidation state of 0 when it is not bonded to any other element;
The O atom in H₂O is bonded to two H atoms. The oxidation state of O in H₂O will be -(2 × (+1)) = -2.

Changes in oxidation states:

S: from -2 to 0.
O: from -1 to -2.

Each S atom will reduce two O atoms. There are two O atoms in each H₂O₂ molecule. There's only one S atom in each H₂S molecule. As a result, pairing one H₂O₂ with one H₂S will balance the change in oxidation state.

Each H₂O₂ contains two O atoms. The two O atoms will produce two H₂O molecules. Each H₂S molecule will lead to one S molecule. Thus the coefficient for H₂O will be two and the coefficient for the rest three species will be one.

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Which two notations represent isotopes of the same element

Mumz [18]

The two notations that represent isotopes of the same element is the one that represented in option 1
The lower number is the number of protons while the upper number is the atomic weight

hope this helps

6 0

3 years ago

Read 2 more answers

What is the theoretical yield of a reaction?

mestny [16]

I think it's D, because theoretical yield is like, the yield you'd get if 100% of the reactants formed to make product. Well that's how I think of it, but it has something to do with limiting reagents and stuff. Sorry this isn't a really detailed explanation.

4 0

3 years ago

What is 1006 in scientific notation?

vladimir1956 [14]

Answer:

1.006 * 103

Explanation:

Add the number between 1 and 9 and add a decimal accordingly . so the answer is 1.006 multiplied by 10 raised to power 3

8 0

3 years ago

g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution

77julia77 [94]

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

<em>Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.</em>

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles <em>(1)</em>

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] <em>(2)</em>

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

8 0

3 years ago