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slega [8]
3 years ago
11

Rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3) in order o

f decreasing concentration of oxygen).
Chemistry
1 answer:
ddd [48]3 years ago
4 0
<span>air = 70 % N2 exhaled air =70%N2 + some less O2 compared to only air 2H2O2=2H2O+O2 NaHCO3=2NaHCO3 → Na2CO3 + H2O + CO2 = No O2 gas from decomposition ssoooo NaHCO3 <exhaled air <air <H2O2

SO sorry for the long wait, hope I helped. 

</span>
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A. Match the term with its definition. (1 point)
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The equation that illustrates the relationship between wave velocity, frequency, and wavelength is v = f×λ.

<h3>Wavelength</h3>

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

<h3>Frequency</h3>

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

<h3>Propagation speed or velocity</h3>

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

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Hope this helps a bit?

6 0
2 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
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