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ser-zykov [4K]
3 years ago
6

Consider this equilibrium for a buffer solution: CH3COOH + H2O CH3COO- + H3O + . When a small amount of acid is added to this sy

stem, which statement is true when equilibrium is reestablished? A. [CH3COO-] and pH have both increased. B. [CH3COOH] and pH have both decreased. C. [CH3COO-] has decreased and pH remains relatively constant. D. [CH3COOH] has decreased and pH remains relatively constant. E. [CH3COO-] has increased and pH remains relatively constant.
Chemistry
1 answer:
Mila [183]3 years ago
6 0

Answer:

e

Explanation:

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Flura [38]
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
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Where is the fulcrum of this lever? circle and label its location on the diagram above
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Answer:

Explanation:

1) A fulcrum is a pivot point that plays a central role (not necessarily located at the center) in a lever. The fulcrum of the attached picture has been circled (in blue).

2) The object placed on this lever's measurement tray is balanced by placing it at the center of the tray. This is the standard way of placing objects on any balance.

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3 years ago
A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

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<span>D. encloses specific structures</span>
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