~Hey guys, can somebody answer this for me, thanks :) ~

Briefly discuss interpretations of your observations and results. Discuss how your observations illustrated LeChatelier's princi

Dvinal [7]

Answer:

Le Chatelier's principle can be applied in explaining the results

Explanation:

According to Le Chatelier's principle, when a constraint such as a change in concentration in this case is imposed on a chemical system in equilibrium, the system will adjust itself in such a way as to annul the constraint imposed.

Hence, when the color of the solution was more like that of the control, the reaction would shift towards the left. Similarly, when the color was more like it was towards the reactant, the reaction would shift towards the right.

If we were to prepare calcium oxalate, we should prepare it in a base solution. This is because when the base was added to calcium oxalate, it did not form any precipitate but when an acid was added to the calcium oxalate, it formed a precipitate.

4 0

3 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

7 0

3 years ago

Determine the energy released per kilogram of fuel used. given mev per reaction, calculate energy in joules per kilogram of reac

tatuchka [14]

The energy released in nuclear reactions are far larger than that released in chemical reactions due to the release of nuclear energy from the nucleus.

<h3>Why is the energy released in a reaction?</h3>

Energy is released in a reaction due to the breaking of bonds are well as formation of bonds.

The quantity of energy released in reactions differs according to the reaction type involved.

When compared to chemical reactions, the energy released in nuclear reactions are very much higher because of the changes that occurs in the nucleus of the atoms involving nuclear energy.

The energy, E released in nuclear reactions is given by the formula below:

$E = mc^{2}$

where m is the mass of the substance and c is the speed of light.

Therefore, the energy released in nuclear reactions are far larger than that released in chemical reactions.

Learn more about nuclear reactions at: brainly.com/question/984564

6 0

2 years ago

Glucose is an example of which carbon-based macromolecule?

Blababa [14]

Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates

Explanation:

carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;

Carbohydrate

Lipids

Proteins and

Nucleic acids.

Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks

8 0

3 years ago

How do the scientist know that cells exist

IrinaVladis [17]

Because of us and other cells from plants and Animals

7 0

3 years ago