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vagabundo [1.1K]
3 years ago
14

Which of the following is a possible set of quantum numbers for an electron n, l, m subscript l, m subscript s? (1, 1, 0, +one o

ver two ) (2, 1, 2, +one over two ) (3, 2, 0, -one over two ) (3, -2, 1, -one over two )
Chemistry
2 answers:
Dovator [93]3 years ago
8 0
The <span>possible set of quantum numbers for an electron n, l, m subscript l, m subscript s is D. 3, -2, 1, -one over two.</span>
Crank3 years ago
7 0

That's incorrect. l can only be between 0 and n-1. m can only be between -l and +l. Your answer is (3,2,0,-1/2)

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Which compounds are classified as Arrhenius acids?
Olenka [21]

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives H^+ ions when dissolved in their aqueous solution.

HA(aq)\rightarrow H^++A^-

Arrhenius bases are those compounds which gives OH^- ions when dissolved in their aqueous solution.

BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

H_2SO_4(aq)\rightarrow 2H^++SO_{4}^{2-}

Hence, the correct answer is option 4.

6 0
4 years ago
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Which substance contains bonds that involve a transfer of electrons from one atom to another
cestrela7 [59]
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3 years ago
If this is a p1000 micropipette, then this is set to dispense [ Select]ul. If this is a p10
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Answer:

1000 µL; 10 µL  

Explanation:

A p1000 micropipet is set to dispense 1000 µL.

A p10 micropipet set to dispense 10 µL.

3 0
4 years ago
Radio waves are ____________ waves​
blsea [12.9K]

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Explanation:

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3 years ago
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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
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Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

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Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

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Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
4 years ago
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