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cricket20 [7]
3 years ago
15

Middle School science 3 1. Illustrate the position of the Earth, moon, and sun during a lunar eclipse. You may do so by typing a

written description or by drawing and labeling a picture.
2. Illustrate the position of the Earth, moon, and sun during a solar eclipse. You may do so by typing a written description or by drawing and labeling a picture.

Chemistry
1 answer:
lilavasa [31]3 years ago
7 0
I hope this helps you

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Which pH corresponds to the equivalence point for the graph below?
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The Ka of hydrazoic acid (HN3) is 1.9 × 10-5 at 25.0 °C. What is the pH of a 0.40 M aqueous solution of HN3?
siniylev [52]

Answer:

The pH of the solution is 2.56.

Explanation:

Given :

Concentration (c) = 0.40 M

Acid dissociation constant = K_a=1.9\times 10^{-5}

The equilibrium reaction for dissociation of HN_3 (weak acid) is,

                           HN_3+H_2O\rightleftharpoons N_3^{-}+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

Dissociation constant is given as:

K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

1.9\times 10^{-5}=\frac{(0.40\alpha)(0.40\alpha)}{0.40(1-\alpha)}

By solving the terms, we get value of (\alpha)}

\alpha=0.00686832

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.4\times 0.00686832=0.002747 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.002747)

pH=2.56

Therefore, the pH of the solution is, 2.56.

8 0
3 years ago
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