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3 years ago

10.Give the possible values for the magnetic quantum number for each of the following orbitals (a) 4p: (b) 3d: (c) 2s (d) 5P

Chemistry

1 answer:

andre [41]3 years ago

5 0

Answer:

(a) 4p: ml = -3, -2, -1, 0, +1, +2, +3

(b) 3d: ml = -2, -1, 0, +1, +2

(c) 2s: ml = -1, 0, +1

(d) 5p: ml = -4, -3, -2, -1, 0, +1, +2, +3, +4

Explanation:

Manetic quantum numbers determine the spatial orientation and number of orbitals in the subshell. ml is from -l to +l.

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Lewis structure for hypochlorus acid

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3 years ago

A Bronsted-Lowry base is defined as a substance that

ICE Princess25 [194]

Answer:

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Explanation

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3 years ago

How do you differentiate between gas and a vapour ?

kolbaska11 [484]

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3 years ago

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

quester [9]

Answer:

$Ksp=1.07x10^{-8}$

Explanation:

Hello,

In this case, the dissociation reaction is:

$PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

For which the equilibrium expression is:

$Ksp=[Pb^{2+}][I^-]^2$

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

$Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M$

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

$[Pb^{2+}]=1.39x10^{-3}M$

$[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M$

Thereby, the solubility product results:

$Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}$

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4 years ago

If you have 6 moles of reactant A and excess of B and C, how much product E would be formed

pychu [463]

B and C are in excess so amount of E will be determined by A.

Amount of product is determined by limiting reagents - Always.

Hence 6 moles of E will be formed.

Hope this helps!

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4 years ago