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kirill115 [55]
3 years ago
8

Write the electron configuration and Noble gas configuration for each element.

Chemistry
1 answer:
cestrela7 [59]3 years ago
7 0

Answer:

1is the answer to that one day F

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What is the net ionic equation for the reaction shown below?
arsen [322]

Answer:

Ag⁺(aq) + Cl⁻(aq)       →      AgCl(s)  

Explanation:

Chemical equation:

AgNO₃(aq) + NaCl(aq)       →     AgCl(s) + NaNO₃(aq)

Balance chemical equation:

AgNO₃(aq) + NaCl(aq)       →     AgCl(s) + NaNO₃(aq)

Ionic equation

Ag⁺(aq)+ NO₃⁻(aq) + Na⁺(aq)+ Cl⁻(aq)       →      AgCl(s)  + Na⁺(aq)+ NO₃⁻(aq)

Net ionic equation:

Ag⁺(aq) + Cl⁻(aq)       →      AgCl(s)  

The NO₃⁻((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The AgCl can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

4 0
3 years ago
1. The vertical columns on the periodic table are called groups, how many groups are
zheka24 [161]

Answer:

18

Explanation:

the numbers at the top tell you

6 0
3 years ago
Read 2 more answers
Pls help 2-3 tyvm ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎
Triss [41]

Answer:

I could I guess?

But like only 1 lol

Explanation:

6 0
3 years ago
g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i
kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is  \alpha_{Y^{-4}} =0.81

So the Conditional Formation constant= K_f =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL  

V e= 25.00 mL  

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.  

[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M

                                                            {K^'}_f

                       Ca²⁺      +      Y⁴          ⇄     CaY²⁻

Initial                 0                  0                      0.001667

change             +x                  +x                     -x

equilibrium        x                    x                    0.001667 - x

{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\

x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M  

pca = —log(2 15x101= 6.7

3 0
4 years ago
Please help :( due tomorrow
olganol [36]
1. 2650000
2. 2410
3. 48100
4 0
2 years ago
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