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MAVERICK [17]
2 years ago
15

The wavelength of the characteristic, bright yellow-orange flame test color of sodium is 590 nm. Calculate the average energy (Δ

E) associated with this atomic emission line.
Chemistry
1 answer:
Assoli18 [71]2 years ago
7 0
<span>The energy associated with the 590nm atomic emission line is 3.369 x 10^49 Joules. This can be determined by multiplying Plank’s constant (6.626 x 10^34 Js) by the speed of light (3x 10^8 m/s), and then dividing the result by the wavelength (590 x 10^-9 m).</span>
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The concept of resonance describes molecular structures Question 17 options: that have several different geometric arrangements.
kipiarov [429]

Answer:

that are formed from hybridized orbitals

Explanation:

The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.  

The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.

3 0
2 years ago
A water molecule is composed of a large oxygen atom bonded to two small hydrogen atoms. Which of the following would make the be
sleet_krkn [62]

The best and most correct answer among the choices provided by your question is the third choice or letter C.

The best model of a water <span>molecule would be: </span><span>Two small, plastic balls attached to a larger plastic ball by toothpicks</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
2 years ago
Read 2 more answers
A solution has a hydroxide-ion concentration of 0.0040 M. What is the pOH of the solution? A solution has a pH value of 3.66. Wh
hodyreva [135]

Answer:

1. pOH = 2.4

2. pOH = 10.34

3. pH = 2.14

Explanation:

1. Determination of the pOH.

Concentration of Hydroxide ion, [OH-] = 0.004 M

pOH =?

pOH = - log [OH-]

pOH = - log (0.004)

pOH = 2.4

Therefore, the pOH of the solution is 2.4

2. Determination of the pOH.

pH = 3.66

pOH =?

pH and pOH are related by the following equation:

pH + pOH = 14

With the above formula, we can obtain the pOH of the solution as follow:

pH = 3.66

pOH =?

pH + pOH = 14

3.66 + pOH = 14

Collect like terms

pOH = 14 - 3.66

pOH = 10.34

Therefore, the pOH of the solution is 10.34

3. Determination of the pH.

Molarity of HCl = 0.0072 M

Concentration of Hydrogen ion, [H+] =?

Thus, we can obtain the concentration of Hydrogen ion, [H+] as follow:

HCl(aq) —> H+(aq) + Cl-(aq)

From the balanced equation above,

1 mole of HCl produced 1 mole of H+.

Therefore, 0.0072 M HCl will also produce 0.0072 M H+.

Therefore, the concentration of Hydrogen ion, [H+] in the solution is 0.0072 M.

Finally, we shall determine the pH of the solution as follow:

Concentration of Hydrogen ion, [H+] = 0.0072 M.

pH =?

pH = - log [H+]

pH = - log (0.0072)

pH = 2.14

Therefore, the pH of the solution is 2.14

3 0
2 years ago
What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi
kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s).

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq).

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

How to rewrite a chemical equation to produce a net ionic equation?

  1. Rewrite all reactants and products that ionizes completely in the solution as ions.
  2. Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

  • Soluble salts,
  • Strong acids, and
  • Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

  • MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as \rm Mg^{2+} and \rm 2\; Cl^{-}.
  • NaOH is a strong base. Each unit of NaOH can be written as \rm Na^{+} and \rm OH^{-}.
  • Mg(OH)₂ is a weak base and should not be written.
  • NaCl is a salt and is soluble in water. Each unit of NaCl can be written as \rm Na^{+} and \rm Cl^{-}.

\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}.

Ions on both sides of the equation:

  • \rm 2\; Cl^{-}, and
  • \rm 2\; Na^{+}.

Add the state symbols:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

For the second reaction:

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}.

\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s).

5 0
2 years ago
Read 2 more answers
Unknown element x has four energy levels, five valence electrons, and is a metalloid. what is elements x?
kykrilka [37]

Answer:

Element X is Arsenic (As)

Explanation:

  • Elements in the periodic table are either metals, non-metals or metalloids.
  • Metals are elements that react by losing electrons to obtain a stable configuration and form cation.
  • Non-metals are those elements that react by gaining electrons to form a stable configuration and form anion.
  • Metalloids are elements in the periodic table that have both metallic and non-metallic properties.
  • Examples of metalloids include Selenium, Arsenic, Boron, etc.
  • Arsenic is a metalloid in period 4 (four energy levels) with five valence electrons.
8 0
2 years ago
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