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Anna71 [15]
3 years ago
8

Where would you find almost no wind at all?

Chemistry
2 answers:
kozerog [31]3 years ago
8 0
Answer:a the prevailing westerlies
suter [353]3 years ago
7 0

Answer:

A. the prevailing westerlies

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Arrange the following element, lithium, potassium, carbon and fluorine in increasing ionization energy
Juli2301 [7.4K]

Answer:

flourine(1681),carbon(1086),lithium(520) and potassium(419)

as we can see that the ionization energy of flourine is the highest than carbon than lithium and than potassium

Explanation:

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3 years ago
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2 years ago
The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con
Aloiza [94]
Use the formula for second order reaction:

\frac{1}{C} = \frac{1}{C_0} + kt

C = concentration at time t 
C0 =  initial conc.
k = rate constant
t = time

1st equation :   \frac{1}{0.56} = \frac{1}{C_0} + 115k

2nd Equation: \frac{1}{0.033} = \frac{1}{C_0} + 735k

Find \frac{1}{C_0} from 1st equation and put it in 2nd equation:

\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k

\frac{1}{0.033} - \frac{1}{0.56} = 620k

k = 0.046
3 0
3 years ago
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

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What is the periodic table
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