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SashulF [63]
3 years ago
11

Complete the acid–base equation for the dissolution of the following compound into liquid HF solvent. The relevant pKa values ar

e given below. Be sure to balance the equations and include the appropriate charges. Phases are optional (liquid, aqueous, etc.).NH3+2HF---->pka in waterHF ---- 3.2NH3 -----38
Chemistry
1 answer:
LenKa [72]3 years ago
7 0

Answer:

The balanced chemical equation: NH₃ + 2 HF → NH₄⁺ + HF₂⁻

Explanation:

According to the Brønsted–Lowry acid–base theory, the acid- base reaction is a type of chemical reaction between the acid and base to give a conjugate acid and a conjugate base.

In this reaction, a Brønsted–Lowry acid loses a proton to form a conjugate base. Whereas, a Brønsted–Lowry base accepts a proton to form a conjugate acid.

Acid + Base ⇌ Conjugate Base + Conjugate Acid

The acid dissociation constant (Kₐ) <em>signifies the acidic strength of a chemical species.</em>

∵ pKₐ = - log Kₐ

Thus for a strong acid, Kₐ value is large and pKₐ value is small.

pKₐ (HF) = 3.2 → strong acid

pKₐ (NH₃) = 38 → weak acid

<u>The chemical reaction involved in the dissolution process:</u>

NH₃ + 2 HF → NH₄⁺ + HF₂⁻

In this acid-base reaction, the acid HF reacts with NH₃ base to give the conjugate base HF₂⁻ and conjugate acid NH₄⁺.

<u>HF (acid) donates a proton to form the conjugate base, HF₂⁻ ion. NH₃ (base) accepts a proton to form the conjugate acid. </u>

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Answer: The mass is 980.6g of Gold.

Explanation:

We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.

Using the Avogadro's number,

6.02 x 10^23 atoms of gold make up 1 mole of gold.

3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.

Now that we know the number of moles, we can then look for the mass using the formular:

Moles = mass/ molar mass

4.98 = mass / 196.9 (atomic mass of gold)

Making "mass" the subject of formula : mass = 4.98 x 196.9= 980.6g

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3 years ago
Which of the following could be true of two species that have a competitve ralationship in the same ecosystems Help me out pleas
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The answer is b. as a whole, the species is mutually beneficial to carry on each others traits and exist in the same ecosystem.
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Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients
Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • First balance the main element in the reaction.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • Now balance oxygen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-+3H_2O

  • Now balance hydrogen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-+6H^+\rightarrow Br^-+3H_2O

  • Now balance the charge.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}+2e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : 3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The balanced chemical equation in acidic medium will be,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

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3 years ago
On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

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