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gogolik [260]
3 years ago
12

What may be expected when k > 1.0?

Chemistry
1 answer:
inessss [21]3 years ago
8 0
When k is greater than 1, this means the equilibrium concentration of the products is greater than the equilibrium concentration of the reactants. K is expressed as the equilibrium concentration of the products divided by the <span>equilibrium concentration of the reactants. This also means the forward reaction is favored. </span>
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mamaluj [8]

Answer:

c

Explanation:

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An igneous rock that contains mostly proxene and olivene has ___.
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It has an ultramafic composition 
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3 years ago
What Is a high-temperature physical state of matter in which atoms lose their electrons?
AlladinOne [14]

Answer:

A plasma.  

Step-by-step explanation:

A <em>plasma i</em>s a hot, ionized gas in which the atoms have lost one or more of their electrons.

Thus, the plasma consists gaseous ions as well as the electrons that were stripped from them.

The Sun's high temperatures strip the electrons from its hydrogen and helium atoms, so the Sun is essentially a giant ball of plasma.

6 0
3 years ago
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
Please help quick<br> Alkali means ___.
sveticcg [70]

Answer:

Metal

Explanation:

In chemistry, an alkali is a basic, ionic salt of an alkali metal or alkaline earth metal chemical element. An alkali also can be defined as a base that dissolves in water. A solution of a soluble base has a pH greater than 7.0.

8 0
3 years ago
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