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jeka57 [31]
3 years ago
13

Write the electron configurations of cu2+ and ag+. is each electron configuration consistent with the color you observed for eac

h cation? explain.
Chemistry
2 answers:
djverab [1.8K]3 years ago
8 0
(1) Copper:
Copper has the atomic number : 29
Cu2+ means that it lost 2 electrons, therefore, the total number of electrons is 29-2 = 27 electrons.
The electronic configuration is:
1s2 2s2 2p6 3s2 3p6 3d9

(2) Silver:
Silver has atomic number : 47
Ag  + means that it lost one electron, therefore, the total number of electrons  is 47-1 = 46 electrons
The electronic configuration is:
<span>1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10</span>

san4es73 [151]3 years ago
8 0

The electron configuration of copper cation:

₂₉Cu²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ or [Ar] 3d⁹.

The electron configuration of copper cation:

₄₇Ag⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ or [Kr] 4d¹⁰.

Copper and silver lose their outer s-orbital electrons before they lose their outer d-orbital electrons.

Each electron configuration is consistent with the color, copper solution is blue, silver solution is colorless.

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Why is it difficult to undergo nucleophilic substitution in haloarene?​
goldenfox [79]

Answer:

In Haloarenes the C atom to which the X group is attached is SP2 hybridized thus it is become difficult to replace it by the Nucleophile. Since arenes and Vinyl halides are electron rich molecules due to presenceof n bonds, they repel Nucleophile attacking them.

8 0
3 years ago
What phase of matter is least common on earth
Diano4ka-milaya [45]
Plasma is the least common matter on earth 
4 0
3 years ago
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mars2501 [29]

Answer:

Qm  = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0
3 years ago
Read 2 more answers
A compound with the empirical formula CH 2 O has a formula mass of 180 g/mol. What is its molecular formula
olga nikolaevna [1]

Empirical formula mass

  • 12+2(1)+16
  • 28+2
  • 30g/mol

Molecular fornula mass:-180g/mol

  • n=Molecular formula mass/Empirical formula mass
  • m=180/30
  • n=6

Molecular formula:-

  • n×Empirical formula
  • 6(CH_2O
  • C_6H_12 O_6
5 0
2 years ago
1. Compare masses: a) 0,4mol CO₂ and 0,6mol H₂O ; b) 0,135mol H₂SO₄ and 0,5mol HCI.
bixtya [17]

Answer:

you can now deduct which one is greater or smaller and by how much.

Explanation:

no of moles= mass/molar mass

1ai) 0.4 = m/ ( 12 + (16*2)

m= 0.4* 44

m= 17.6g

ii) 0.6= m/( 2*1 + 16)

m= 0.6 *18

m= 10.8g

b) 0.135 = m/ ( 2*1 +32 + (16*4)

m= 0.135* 98

m= 13.23g

ii) 0.5= m/ (1+35.5)

m= 0.5*36.5

m= 18.25g

2. Avogadro's Number = 6.02×10²³

1 mol of any element= 6.02×10²³ particles

a) 0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules

ii) 0.3 mol of CO2= (6.02×10²³) * 0.3

= 1.806 × 10²³ molecules

Ans: 0.3 mol of CO2

bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

= 1.505 × 10²³ molecules

bii) - find the no of moles first:

no of moles= mass/molar mass

n= 3.4g/ 34g →mr of H2S in g=2+32= 34g

n= 0.l mol

- use the Avogadro Number.

0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules.

biii) here you're given the density, use it to find the mass of acetic acid.

ρ = 1049 g/ml

ρ = m/v, where v=5 ml

1049 = m/ 5

m= ρ*v

m= 1049*5

m= 5245 g

• convert this into moles.

mr of CH3COOH= 12 + 3+ 12+ 16+ 16+ 1

= 60

mr in g = 60g

n= m/mr

n= 5245/ 60

n= 87. 41666...

n= 87.4 moles

•using Avogadro's Number:

87.4 moles of acetic acid=(6.02×10²³)*87.4

= 2.25148* 10²⁵

= 2.25 * 10²⁵ molecules

thus, the ans for this is 5 ml of acetic acid.

7 0
2 years ago
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