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dedylja [7]
3 years ago
9

What would happen if a TLC plate was placed in a jar where the solvent level was above the level of the origin (sample spot)? Wh

at should you do if you inadvertently spotted your plate too low (aside from consulting your TA)?
Chemistry
2 answers:
Hitman42 [59]3 years ago
8 0

Answer:

Thin layer chromatography (TLC) is a form of qualitative analysis called adsorption

chromatography used almost daily in synthetic organic chemistry labs. The composition of a

mixture can be assessed relative to known standards and separation of the mixture is roughly

based on polarity. In brief, a small portion of the sample is placed on the stationary phase

(polar silica) and is carried up the TLC plate by the mobile phase (solvent). A polar compound

will interact more with the polar stationary phase and will not travel as far up the plate.

Conversely, a less polar compound will move farther up the plate. Notice that this technique is

not terribly different from GC. Separation is based on polarity in both cases – relative boiling

points can be estimated from polarity. Mobile and stationary phases are still employed, except

that the states of matter are different. The principles of TLC will later be applied to column

chromatography, where larger sample volumes will be physically separated and isolated based

on polarity of the components. In this lab, students will use solvents of different polarity (mobile

phases) to determine the optimal separation of the pigments found in spinach extracts.

The most prominent types of plant pigments are chlorophylls, carotenoids, flavanoids,

and tannins. Chlorophylls contain a ring system formed by four pyrroles linked by four methine

bridges, a Mg2+ ion in its center, and a long nonpolar hydrocarbon chain (C20H39). Chlorophylls are

cousins of other biologically important molecules such as vitamin B12 and the heme found in

hemoglobin; they all have a tetrapyrrole ring system. There are two main types of chlorophylls

present in higher plants, a and b. Chlorophyll a is more abundant than chlorophyll b by a ratio of 3

to 1. The only structural difference between them is that a methyl group in a has been replaced by

a formyl group, CHO, in b.

Explanation:

nika2105 [10]3 years ago
7 0

Answer: ok

Explanation:

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A compound contains 72% magnesium and 28% nitrogen. What is its empirical formula?
SashulF [63]

So the empirical formula is Mg3N2

7 0
3 years ago
Iron reacts with cl2 according to the equation 2 fe(s) + 3 cl2(g) ? 2 fecl3(s). how many moles of cl2 is needed to react with 4.
IceJOKER [234]
2Fe + 3Cl₂ ---> 2FeCl₃

4.4mol of Fe, you have a 2:3 ratio of Fe to Cl₂ so divide 4.4/2 = 2.2 and multiply by three 2.2 x 3 = 6.6mol of Cl₂

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8 0
3 years ago
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To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?
evablogger [386]

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is \rm  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen  

\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

brainly.com/question/26416088

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3 0
2 years ago
Balance :FeCl3 + __Ca(OH)2 → ___ Fe(OH)3 + __ CaCl2
ArbitrLikvidat [17]
Your answer is 2:3:2:3
7 0
3 years ago
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


7 0
3 years ago
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