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Mamont248 [21]
3 years ago
12

Select the procedure that will NOT help saturate a thin‑layer chromatography (TLC) developing chamber with developing solvent va

pors.
(A) Cover the chamber.
(B) Gently swirl the solvent in the chamber prior to placing the TLC plate inside.
(C) Add ninhydrin (a visualization technique) to the developing solvent.
(D) Place a paper wick (like a piece of filter paper) inside the chamber.
Chemistry
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

C)

Explanation:

(A) Cover the chamber. => <u>It helps to keep the solvent in his gas state</u>

<u />

(B) Gently swirl the solvent in the chamber prior to placing the TLC plate inside => <u>The mechanical energy can promote the conversion from the liquid state to the gas state and help to the saturation process.</u>

<u></u>

(C) Add ninhydrin (a visualization technique) to the developing solvent =><u>There is no change in the saturation process</u>

(D) Place a paper wick (like a piece of filter paper) inside the chamber=><u>Increases the area for evaporation. The solvent can go up across the paper by capillarity a then can be evaporated increasing the saturation in the chamber.</u>

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The number is written in scientific notation.


The significant figures are the numbers known precisely plus 1 digit that is uncertain.


When the numbers are expressed in scientific notation all the digits before  10^n are significant.


So, 7 and 8 are significant figures and the answer is that the number of significant figures is 2.


Answer: 2
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Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
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To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


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3 years ago
Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.
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