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DedPeter [7]
2 years ago
8

How did dmitri mendeleev arrange the periodic table.

Chemistry
1 answer:
worty [1.4K]2 years ago
3 0

Answer:

order of increasing relative atomic mass

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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:
Lubov Fominskaja [6]

Answer:

\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}

Explanation:

(a) Balanced equation

Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag

(b) Calculation

You want to convert atoms of Cu to atoms of Ag.

The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}

4 0
3 years ago
Read 2 more answers
How many moles are there in 87.2 g of zinc fluoride?
Sladkaya [172]

Answer:

what I got was 0.8435160945347224 moles

7 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
What is acid according to arhenias concept?
nignag [31]

Answer:

Acid are those substances which release H + ions when dissolved in water.

Get that hundooo!

7 0
3 years ago
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