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3 years ago

Consider the redox reaction below.

Chemistry

1 answer:

vovangra [49]3 years ago

4 0

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

Oxidation (loss of electrons or increase in the oxidation state of entity)
Reduction (gain of electrons or decrease in the oxidation state of the entity)

An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

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A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

, Hydrocarbons are compounds that contain(1) carbon, only(2) carbon and hydrogen, only(3) carbon, hydrogen, and oxygen, only(4)

Firdavs [7]

Hydrocarbons are compounds formed by only hydrogen atoms and carbon.

Answer (2)

hope this helps!

4 0

3 years ago

Occupational Safety and Health Administration has set the limit on the maximum percentage of carbon dioxide (CO2) in air a worke

Lyrx [107]

Answer:

The minimum rate of fresh air in the room is 176 moles/min

Explanation:

High exposure of CO₂ has health effects as headaches, increased heart rate, elevated blood pressure, coma, asphyxia, convulsions, etc.

0,500 mole% of CO₂ in air means 0,500 moles of CO₂ per 100 moles of air

As the rate of sublimation of CO₂ is 0,880, the minimum rate of fresh air in the room must be:

$\frac{0,500 moles CO_{2}}{100 moles Air} = \frac{0,880 moles CO_{2}/min}{X}$

X = 176 moles of Air/min

I hope it helps!

4 0

3 years ago

What would a 1:1 liquid binary mixture of Cyclohexane and Toluene look like? How would the molecules associate with one another?

Ainat [17]

Answer:

Both compounds are colorless, so a mixture of them will look transparent. Both molecules are nopolar, therefore they will associate with each other through London interactions, also known as Van del Waals forces or as transient dipole-transient dipole interactions.

Explanation:

3 0

3 years ago

The processof watermoving on the earth surface andin the atmosphere is call

vfiekz [6]

This process is called the water cycle

5 0

3 years ago

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