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Sholpan [36]
3 years ago
10

Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolat

e and measure the amount of gas produced.
Suppose the CO gas evolved by a certain chemical reaction taking place at 55 degree C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 117 ml.

Calculate the mass of CO that is in the collection tube. Round your answer to 2 significant digits. You can make any normal and reasonable assumption about the reaction conditions and the nature of the gases.
Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Answer:

The correct answer is 0.12 grams.

Explanation:

The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.

Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.  

The number of moles or n can be determined by using the equation, mass/molar mass.  

R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).  

On putting the values we get:

n = PV/RT

= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)

= 0.0043447 mol

Therefore, mass of CO will be moles * molar mass of CO

= 0.0043447 mol * 28 g/mol

= 0.12 g

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Alik [6]
<span>Correct answer is:

C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy

But how to get there?

Let's start with simple explanation of what exactly is cellular respiration. 

Cellular respiration is a multistage biochemical oxidation process of organic substances when prime product is energy (ATP - adenosine triphosphate) and other are released waste products. Cellular respiration takes place even if other metabolic processes are stopped, but cellular respiration may differ in particular organism groups.Some reactions during whole process of cellular respiration are similar in all types of living organisms.

Cellular respiration is prime indication of declining living processes.Only viruses which are on the edge of living organism and chemical particle are not performing cellular respiration.But to the point :P

In cellular respiration all substrates which are in the cell might be organic, but mostly we are using sugar oxidation - glucose in the presence of oxygen. Chemical formula of sugar looks like this:

C_{6}H_{12}O_6

Oxygen is just

O_2

so for now we have just part of the equation:

C_{6}H_{12}O_6+O_{2}\Rightarrow

But what would be on the right hand side?

It's quite simple, remember equation of full combustion? If we want to burn something we need oxygen like in the equation, so the product of this equation would be carbon dioxide, water and of course energy (ATP).Carbon dioxide formula looks like this:

CO_{2}

As a reminder water formula:

H_{2}O

Full formula would look like that:

C_{6}H_{12}O_6+O_{2}\Rightarrow\ CO_{2}+H_{2}O+energy

But still as you see this equation is unbalanced, after balancing it would like that:

C_{6}H_{12}O_6+6O_{2}\Rightarrow 6CO_{2}+6H_{2}O+energy

At the end I would like to explain one more thing. Energy which has been released during this process is part of high-energy connection which might be used to perform chemical reactions in the cell or to move organism for example in muscles. We need to remember that production of ATP is not happening with 100% efficiency and part of this energy is released as heat.</span>
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10) One mole of a diatomic ideal gas is initially at a temperature of 127 °C and
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8 0
2 years ago
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et
Alexxx [7]

<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For testosterone:</u>

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol

  • <u>For diethyl ether:</u>

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol

Mole fraction of a substance is calculated by using the equation:

\chi_A=\frac{n_A}{n_A+n_B}

\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}

\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095

The formula for relative lowering of vapor pressure will be:

\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}

where,

p^o = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

p^s = vapor pressure of the solution = ?

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

\chi_{\text{solute}} = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg

Hence, the vapor pressure of solution is 459.17 mmHg

7 0
3 years ago
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Write, in ink, either sample or blank on the side of the cuvette to keep track of them. This is so since sample and blank is not absorbed at the same time by the machine.

7 0
3 years ago
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