Can someone give me brief background info about chromatography experiment?

Answers

Answer 1
Answer:

Chromatography is a pretty accurate description of what happens to ink on wet paper, because it literally means "color writing" (from the Greek words chroma and graphe). Really, though, it's a bit of a misnomer because it often doesn't involve color, paper, ink, or writing. Chromatography is actually a way of separating out a mixture of chemicals, which are in gas or liquid form, by letting them creep slowly past another substance, which is typically a liquid or solid. So, with the ink and paper trick for example, we have a liquid (the ink) dissolved in water or another solvent creeping over the surface of a solid (the paper).

The essential thing about chromatography is that we have some mixture in one state of matter (something like a gas or liquid) moving over the surface of something else in another state of matter (a liquid or solid) that stays where it is. The moving substance is called the mobile phase and the substance that stays put is the stationary phase. As the mobile phase moves, it separates out into its components on the stationary phase. We can then identify them one by one.


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Refer to the following balance equation : 2c2h6 + 7o2 = 4co2 + 6 h20. How many moles of c2h6 will combust completely with 2.0 moles of oxygen gas. How many grams of water will be produced when 30.0 g of c2h6 reacts completely with oxygen?

Answers

1. From the balanced equation given above, the ratio of the number of moles of the hydrocarbon and oxygen is equal to 2/7. Given that there are 2 moles of oxygen,

    moles hydrocarbon = (2 moles O2)(2 moles HC / 7 moles oxygen)

Simplifying,
   moles HC = 4/7 or 0.57 moles HC

Answer: 0.57 moles HC

2. The calculation for the mass of water is shown below with the dimensional analysis and conversion factors,

  (30 g C2H6)1 mole C2H6/30 g C2H6)(6 molesH2O/2 moles C2H6)(18 g H2O/1 mole C2H6)


Simplifying,
     mass  = 54 grams of water

How many electrons are in Fe3+ ?

Answers

23 electrons hope this helps

Answer:

There are 23 electrons in Fe3+

Classify each of the following as a compound, a homogeneous mixture, or a heterogeneous mixture: (a) orange juice; (b) vegetable soup; (c) cement; (d) calcium sulfate; (e) teas

Answers

Explanation:

A compound is a substance that contains more than two different atoms in a fixed ratio by mass. Elements of a compound are chemically combined together.

A homogeneous mixture is defined as a mixture in which solute particles are uniformly distributed into the solvent.

Whereas a heterogeneous mixture is defined as a mixture in which solute particles are non-uniformly distributed in a solvent.

In both homogeneous and heterogeneous mixture there is no chemical combination between solute and solvent particles.

Therefore, each given substance is classified as follows.

(a) Orange juice :  As it contains pulp of orange along with the liquid which are not present in uniform composition. Hence, orange juice is a heterogeneous mixture.

(b) vegetable soup : This soup also contains vegetable pieces and spices which are non-uniformly distributed. Hence, it is also a heterogeneous mixture.

(c) cement : It contains several different compound present in definite composition. Hence, cement is a homogeneous mixture.

(d) calcium sulfate : Its chemical formula is CaSO_(4) and in this compound both calcium and sulfate atoms are chemically combined together. Therefore, calcium sulfate is a compound.

(e) Tea : It is a homogeneous mixture as there is uniform composition of tea water in milk.

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half reaction, and which side they appear on. Si (s) + Mg(OH)2 (s) → Mg (s) + SiO32- (aq)

Answers

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

A rigid container of gas has a pressure of 1.72 atm and temperature of 21 oC. If the temperature increases to 85 oC, what is the new pressure in atm?

Answers

Answer:

the new pressure is 2.09 atm

Explanation:

you have to use gay lussac's law so the formula is

p1/t1 = p2/t2

and convert C to Kelvin k=C+273.15

1.72atm/294.15 = p2/358.15

solve for p2 by multiplying 358.15 on both sides

p2=2.09 atm

A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answers

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial  = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work