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3 years ago

What is a radical in science

1 answer:

7 0

Radical In chemistry, radicals are atomic or molecular species with unpaired electrons on an otherwise open shell configuration. These unpaired electrons are usually highly reactive, so radicals are likely to take part in chemical reactions.
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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /

barxatty [35]

Answer:

$[Pb^{2+}]=3.9 \times 10^{-2}M$

this is the concentration required to initiate precipitation

Explanation:

$PbCl_2$ ⇄ $Pb^{2+}+2Cl^-$

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

$Ip=[Pb^{2}][2Cl^-]^2=Ksp$

$Ksp=2.4\times 10^{-4}$

lets solubility=S

$[Pb^{2+}] = S$

$[Cl^-]=2S$

$Ksp=[Pb^{2+}]\times [Cl^-]^2$

$Ksp=S \times (2S)^2$

$Ksp=4S^3$

$S=\sqrt[3]{\frac{Ksp}{4} }$

$S=3.9\times 10^{-2}$

$[Pb^{2+}]=3.9 \times 10^{-2}M$ this is the concentration required to initiate precipitation

4 0

3 years ago

How to revise in just 1 whole day for chemistry combined science is it possible to get a good grade

Pavlova-9 [17]

it will be hard, but you can do it. Just study given the materials for the course. Understand enthalpy and entropy, and various types of bonding and you'll be fine.

4 0

3 years ago

Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

ki77a [65]

<span>

</span> $Cu_4(OH)_6(SO_4)$
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = $\frac{2}{3} \times 2.3 \times 10^{-6}$

$= 1.53 \times 10^{-6} 
$

6 0

4 years ago

(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would

Lelu [443]

Answer:

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0

3 years ago

Plss help me!!!!!!!!

Brums [2.3K]

Answer:

rise,cool..................

8 0

3 years ago

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