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dezoksy [38]
3 years ago
7

PLEASANSWERSOONThe elements on the Periodic Table that we use today are arranged by

Chemistry
1 answer:
OlgaM077 [116]3 years ago
4 0
The answer is atomic number.
Explanation: Meendelev believed in putting it in order by atomic mass but that didn’t make since so another scientist by the name of Henry Moseley arranged the elements by atomic number which made the periodic table make more since. Because they were in order of which properties they had.
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In an aqueous solution of a certain acid the acid is 0.10% dissociated and the pH is 4.06. Calculate the acid dissociation const
Kruka [31]

Answer : The acid dissociation constant Ka of the acid is, 8.7\times 10^{-5}

Explanation :

First we have to calculate the concentration of hydrogen ion.

pH=-\log [H^+]

Given: pH = 4.06

4.06=-\log [H^+]

[H^+]=8.71\times 10^{-5}M

The dissociation of acid reaction is:

                         HA\rightarrow H^++A^-

Initial conc.      c        0         0

At eqm.           c-cα    cα       cα

Given:

Degree of dissociation = α = 0.10 % = 0.001

[H^+]=c\alpha

8.71\times 10^{-5}=c\times 0.001

c=0.0871M

The expression of dissociation constant of acid is:

K_a=\frac{[H^+][A^-]}{[HA]}

K_a=\frac{(c\times \alpha)\times (c\times \alpha)}{(c-c\alpha)}

Now put all the given values in this expression, we get:

K_a=\frac{(0.0871\times 0.001)\times (0.0871\times 0.001)}{(0.0871-0.0871\times 0.001)}

K_a=8.7\times 10^{-5}

Thus, the acid dissociation constant Ka of the acid is, 8.7\times 10^{-5}

7 0
3 years ago
A(n) __________ eclipse might occur when the Earth, Moon, and Sun are in the positions shown in the picture.
zheka24 [161]

Answer:

The answer is most likely D

3 0
3 years ago
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A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of
zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

  • CH3COOH ↔ CH3COO-  +  H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ <em>C</em> CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

  • CH3COOH + KOH ↔ CH3COONa + H2O

∴ <em>C </em>CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ <em>C</em> CH3COOH = 0.05 M

∴ <em>C</em> KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

  • (<em>C</em>*V)acid = (<em>C</em>*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ <em>C</em> CH3COOH = 0.0143 M

∴ <em>C</em> KOH =  0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

  • CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = <em>C</em> CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ <em>C</em> KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ <em>C</em> KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0
3 years ago
What is the specific heat of copper when heated to 221.32
Dafna1 [17]

Answer:

The specific heat of copper when heated to 221.32 (not listed form of heat measurement) is 221.32 (not listed form of heat measurement).

Explanation:

uh not really sure what else there is here, I may be missing something

5 0
2 years ago
What does conserving mass mean in a chemical equation?
lubasha [3.4K]
The law of conservation of mass states that mass or matter cannot be created or destroyed, only transferred or recombined.
For chemical equations, this law means that each element must be accounted for equally both for reactants and products. So the same numbers of each atom must match on each side, hence the necessity for balancing the chemical equation accurately. This created a field of chemistry called Stoichiometry, which accounts for the conservation of matter throughout chemical reactions and processes.
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