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Dmitry [639]
2 years ago
14

If there was another spot on the TLC plate, what would be that other spot assuming all starting material was consumed? at what s

tage of the workup do we separate this biproduct from triphenylmethanol?
Chemistry
1 answer:
Assoli18 [71]2 years ago
7 0

The other spot that would be assuming all starting material was consumed are drop or splash an organic compound on the plate.

When we pour a petroleum on the solid mixture. It will mixed the biphenyl but not the triphenylmethanol and this causes separation.

The full form of TLC Plate is Thin Layer Chromatography Plates

It is a ready-made plates used which are chemically fixed.

The halted phase is draw on on its surface in the form of a thin layer. The halted phase on the plate has a fine particle size and also has a consistent thickness.

Uses of TLC Plates : for estimating the purity in the chemical formula and for finding compounds.

It is a composition of silica gel with a back support of plastic.

It shows UV absorption.

To know more about TLC Plates here :

brainly.com/question/17562109?referrer=searchResults

#SPJ4

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artcher [175]
Ca + ZnCO3 → CaCO3 + Zn
3 0
3 years ago
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What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?
Viefleur [7K]
Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³ 
y =  \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}
\boxed{y \approx 0.04\:mol}


Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

x = 40*0.04
\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark

Answer:
The mass is 1.6 grams
6 0
3 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

<em>= 14, 508 J/K/kg</em>

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

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