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galina1969 [7]
2 years ago
11

When was that one moment when you laughed the hardest

Chemistry
2 answers:
kolezko [41]2 years ago
7 0

Answer:

last month!!

it was on yt

Andrei [34K]2 years ago
4 0

Answer:

My weightlifting coaches are all gay for each other and its hilarious to see 3 large buff men play fighting and wrestling around the weight room.

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Part C<br> How did Dr. Tierno find the answer to his question?
matrenka [14]
Sorry I cant I just need some points
7 0
3 years ago
Read 2 more answers
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

5 0
2 years ago
These questions are from an experiment where we had a mixture of Ferrocene, acetylferrocene and diacetyl ferrocene and we separa
PilotLPTM [1.2K]

Answer:

Explanation:

The polarity of the 3 compounds would be in the order of

Ferrocene < Acetylferrocene < Diacetylferrocene

Your TLC data has to also support this observation . This can be checked by measuring the values of Rf ( Retention factor = distance travelled by solute/solvent ) .The Rf values also has to follow this particular order: -

Ferrocene > acetylferrocene > diacetylferrocene

2) Hexane happens to be a non-polar solvent. The polarity of hexane can be increased if some polar solvents for example, ethyl and methylene chloride etc are added

Therefore, in the increasing order of solvents polarity, we have

Hexane < 1:1 mixture of hexane: methylene chloride < 9:1 mixture of methylene chloride:

3) Chromatographic techniques all have a stationary phase in addition to a mobile phase. In the case of column chromatography, the silica gel will be the stationary phase and the solvent that will be poured will be the mobile phase.

4) The TLC and column chromatography both happen to have the same stationary phase which is the silica gel. Also, the same solvent mixture is used in both the techniques. This makes the result of the 2 to be almost the same. The difference seen between them is that, TLC works against the gravity while on the other hand column chromatography works in the direction of the gravity.

5) The key feature in the IR spectra of the acetylferrocene that will be absent in the spectra of ferrocene is the presence of carbonyl stretching frequency at close to 1700 per cm(cm-1). This peak is easily differentiated between both acetyl ferrocene and ferrocene.

4 0
2 years ago
How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat
stira [4]

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

8 0
3 years ago
What is the shape of a molecule containing three covalent bonds with one unshared pair?
Schach [20]

pyramidal, if you have a molecule kit, i would strongly recommend watching a video of someone using a kit explaining the different shapes and following along. It helps a lot!

7 0
3 years ago
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