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Vesna [10]
2 years ago
15

Um Is it in third period and has three electrons in its highest P orbitals

Chemistry
1 answer:
anyanavicka [17]2 years ago
7 0

Answer:

Shut yo musty dusty ah up zaddy

Explanation:

come here big mami

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Transamination removes the amino group to form a carbon skeleton that contains
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k

Explanation:

kidfkvyg bt tgk mg RH g fu f

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3 years ago
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Describe the stages of volcanic activity
OverLord2011 [107]
Active, Dormant, and Extinct.
Hope this helps.
5 0
2 years ago
Write the word and balanced chemical equations for the reaction between:
nadezda [96]

Answer:

3 HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

Explanation:

An acid reacts with a base producing water and a salt. Having this in mind the reaction of nitric acid (HNO₃) and Iron (III) hydroxide (Fe(OH)₃) is:

HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

<em>The H⁺ of the acid reacts with the OH⁻ to produce H₂O. The other ions (Fe³⁺ and NO₃⁻) produce the salt</em>

<em />

There are 3 nitrates in products. To balance the nitrates:

<h3>3 HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃</h3>

<em>And this is the balanced reaction</em>

<em />

3 0
3 years ago
Para el elemento de Z-56 b, que formula minima tiene el compuesto que forma con 17-cl¿ c, que tipo de union hay en dicho compues
inessss [21]

Answer:

ver explicacion

Explanation:

El elemento que tiene el número atómico 56 en la tabla periódica del bario. El bario es un elemento del grupo dos.

Se combina con el cloro para formar cloruro de bario. que tiene la fórmula BaCl2 de acuerdo con la valencia de ambos elementos.

El cloruro de bario es un compuesto iónico.

5 0
3 years ago
If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T
Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

         7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔      log([Conjugate base]/[Acid])  = 0

⇔     [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

                ⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

                ⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

    = 7.55 + log(0.003 / 0.009) = 7.07

8 0
2 years ago
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