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4 years ago

Write the symbol for every chemical element that has atomic number greater than 7 and atomic mass less than 24.0 u.

1 answer:

7 0

<span>This question can best be answered using a Periodic Table of Elements. All the elements are listed by their atomic number starting from the top left and moving right and down the table. Starting at element 8, you will find Oxygen with a mass of 15.99 amu, followed by element 9 Fluorine with a mass of 18.99 , element 10 Neon with a mass of 20.18 , element 11 Sodium with a mass of 22.99. Writing them with their symbols you would have; O, F, Ne, Na.</span>

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What is the oxidation number of hydrogen in each of the following species?

Keith_Richards [23]

Explanation:

d. H = -1

a. H = +1

b. H= +1

c. H = +1

e. H= 0

f. H= +1

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3 years ago

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Answer:

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Explanation:

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3 years ago

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3 years ago

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3 years ago

Calculate the [OH-] and the pH for a solution of 0.24M methylamine, CH3NH2. Kb = 3.7 X 10-4.

Naddika [18.5K]

Answer:

$[OH^-]=9.24x10^{-3}M$ .

$pH=11.97$ .

Explanation:

Hello,

In this case, since the ionization of methylamine is:

$CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)$

The equilibrium expression is:

$Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

And in terms of the reaction extent $x$ which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:

$3.7x10^{-4}=\frac{x*x}{024-x}$

Whose solution for $x$ via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:

$[OH^-]=x=9.24x10^{-3}M$

With which we can compute the pOH at first:

$pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034$

Then, since pH and pOH are related via:

$pH+pOH=14$

The pH turns out:

$pH=14-pOH=14-2.034\\\\pH=11.97$

Best regards.

8 0

3 years ago