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3 years ago

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In the stratosphere, oxygen (o2) is converted to ozone (o3) by high energy ultraviolet light. when the equation is written and b

alanced using whole number coefficients, the coefficient of ozone is 1. 4. 2. 3.

1 answer:

Sergio039 [100]3 years ago

8 0

3O₂ = 2O₃ (ozone), so coefficient is 2.

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Explain that : how we explore different between Licl and Nacl

Leona [35]

Answer:

See the explanation below, please.

Explanation:

In the bunsen burner, the gas and air inlet can be regulated manually. In the case of metals (such as lithium and sodium in this case) they contain an electron in the latter in its external electronic configuration. They are characterized by transferring electrons easily and produce the emission of light when excited, producing flames of different colors in the lighter (orange for sodium and red / scarlet for lithium)

8 0

3 years ago

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

<u>Answer:</u> The molecular formula of the compound is $C_4H_{10}O_4$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

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

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

<u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound 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 $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

<u>Step 2:</u> 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 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

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 = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

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

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot

barxatty [35]

Ethanoic acid ionizes in aqueous solutions to form two ions which are $CH_3COO^-$ and $H^+$

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula $CH_3COOH$ .

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

$CH_3COOH ---- > CH_3COO^- + H^+$

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single $H^+$

More on ethanoic acid can be found here: brainly.com/question/9991017

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8 0

2 years ago

A non-magnetized metal object can be magnetized with a stronger magnet. Explain how this happens on the atomic level.

sergeinik [125]

I believe the atomic level increases hope this helps.

6 0

3 years ago

Which expression can be used to calculate the density of a 129.5-gram sample of bronze that has a volume of 14.8 cubic centimete

PilotLPTM [1.2K]

Answer:

it would be like 30

Explanation:

just divide the gram by cubic

5 0

2 years ago