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3 years ago

5

Why does the boiling point of water change at high elevation (maybe in Denver, Colorado, the Mile-High City) compared to sea lev

el (maybe in Ocean County, New Jersey)?

1 answer:

Oksanka [162]3 years ago

7 0

Explanation:

This is due to a reduction in atmospheric pressure – pressure due to the weight of the overlying air column). Remember that the higher the altitude the lower the air density and subsequently the lower the air pressure.

Therefore when heating water at higher altitude the vapor pressure of the water at the surface overpowers the atmospheric pressure – which is slightly lower than at sea level – even before the water temperatures reach 100 degrees centigrade.

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Which model is based on Bohr's quantum model?

Kaylis [27]

Answer:

Model D

Explanation:

Bohr's Model has a planetary look. Where the electrons are in an orbit.

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3 years ago

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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 $C_4H_{10}O_4$

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3 years ago

What is the difference between chemical and physical change with respect

kozerog [31]

Answer:

Physical changes are changes that do not alter the identity of a substance. Chemical changes are changes that occur when one substance is turned into another substance.

Explanation:

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What is some fun facts about the person who started Sand Storms? What can be the worst in a Sand Storms? If you were caught in o

Dahasolnce [82]

Answer:

The clouds that carry the dust can be huge, and miles long. They can rise to be over 305m. They also have wind speeds of at least 40 kilometers.Sandstorms can happen really quickly.

Explanation:

Sandstorms carry with them large volumes of sand and dust. But more than sand, they also carry with them virus spores that interact with the atmosphere. Because of this, I can contact any disease from it.

Other immediate effects of sandstorms to me include the worsening of lung functions if I have asthma. Due to inhalation of large amounts of dust, dust pneumonia could be developed. Diseases such as silicosis could also develop from prolonged exposure to sand. If left untreated, it will eventually lead to asphyxiation and lung cancer. The eyes could also be affected by the particles. Keratoconjunctivitis sicca, or dry eyes, could develop which could lead to blindness. My exposure to sandstorms could have adverse effects on my circulatory system. So I could either die immediately or fall terrible sick it depend on my health issue.

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3 years ago

You bought a new fish aquarium with the dimensions of 55 cm x 100 cm x 80 cm. what volume of water should you put in it? Show yo

harina [27]

Since its volume you do what it says 55 cm x 100 cm x 80cm and do 100 x 50 which is 5,000. then 5,000 x 80 which is 40,000. sorry i cant show work on here.

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4 years ago