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spayn [35]
3 years ago
9

What could be an electron configuration of Na in the excited state?

Chemistry
1 answer:
olasank [31]3 years ago
4 0

Answer:

The excited state electron configuration shows when an electron is excited and jumps into a higher orbital. For example, sodium in its excited state would have an electron configuration of 1s2 2s2 2p6 3p1, compared with its ground state of 1s2 2s2 2p6 3s1.

Explanation:

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Answer this question fast please!
MrRissso [65]

Answer:

A is a physical change, no atoms are being lost its still in it original form

B is a chemical change, it is losing atoms and changing into a new substance

Explanation:

5 0
3 years ago
Which of the following demonstrates a poor internal control procedure?
Alex_Xolod [135]

Answer: The bookkeeper makes cash deposits and records journal entries related to cash, while the treasurer prepares the bank reconciliation.

Explanation: The bookkeeper in an organization is the person in charge of records keeping, his basic task is to keep financial records and therefore the bookkeeper has no business handling physical cash in an organization.

The financial secretary on the hand is in charge saving cash income and withdrawal of cash from an organization, this is done with the organization's accountant approval.

5 0
3 years ago
A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r
EastWind [94]
M_{Al}=26,98\frac{g}{mol}\\
m=51g\\\\
n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\
M_{NaOH}=39,4\frac{g}{mol}\\
m=84,1g\\\\
n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

M_{NaOH}=39,4\frac{g}{mol}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
n=2,835mol\\&#10;M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g

7 0
3 years ago
In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of wate
maxonik [38]

Answer:

21.344%

Explanation:

For the given chemical reaction, 8 moles of the reactant should produce 4 moles of Na_{2}O. However, 195 g of Na_{2}O was produced instead. The molar mass of Na_{2}O is 61.9789 g/mol.

Thus, the moles of Na_{2}O produced = 195/61.9789 = 3.1462 moles

The percent error = [(Actual -Experiment)/Actual]*100%

The percent error =  [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%

8 0
4 years ago
Given the balanced equation representing a reaction:
Vilka [71]
(2) a base because they accept H+ ions. NH3 is the conjugate base of NH4+.
5 0
4 years ago
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