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Lesechka [4]
4 years ago
15

Rank the solutions in order of decreasing [H3O ]. Rank solutions from largest to smallest hydronium ion concentration. To rank i

tems as equivalent, overlap them. a. 0.10 M HNO3.
b. 0.10 M HCN.
c. 0.10 M HNO2.
d. 0.10 M HClO.e. smallest concentration.f. largest concentration.
Chemistry
1 answer:
Usimov [2.4K]4 years ago
5 0

Answer:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

Explanation:

We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.

HNO₃ is a strong acid and will have the largest hydronium concentration.

HCN  Ka = 6.2 x 10⁻¹⁰

HNO₂ Ka = 4.0 x 10⁻⁴

HClO  Ka = 3.0 x 10⁻⁸

The ranking from smallest to largest hydronium concentration will then be:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

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Strong acids are assumed 100% dissociated in water. True As a solution becomes more basic, the pOH of the solution increases. Fa
Kruka [31]

Answer:

Strong acids are assumed 100% dissociated in water- True

As a solution becomes more basic, the pOH of the solution increases- false

The conjugate base of a weak acid is a strong base- true

The Ka equilibrium constant always refers to the reaction of an acid with water to produce the conjugate base of the acid and the hydronium ion- True

As the Kb value for a base increases, base strength increases- true

The weaker the acid, the stronger the conjugate base- true

Explanation:

An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.

The pOH decreases as a solution becomes more basic (as OH^- concentration increases).

Ka refers to the dissociation of an acid HA into H3O^+ and A^-.

The greater the base dissociation constant, the greater the base strength.

The weaker an acid is, the stronger , its conjugate base will be.

7 0
3 years ago
If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
3 years ago
Help with this chemistry worksheet
viktelen [127]

Answer:

1. is exothermic

Explanation:

5 0
3 years ago
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3 0
3 years ago
Which of the four cars had the least net force applied to it when it was launched down the ramp
castortr0y [4]

The least net force applied : Car 3(12 N)

<h3>Further explanation  </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

Car 1 ⇒m=0.5 kg, a=36 m/s²

\tt \sum F=0.5\times 36=18~N

Car 2⇒m=0.8 kg, a=50 m/s²

\tt \sum F=0.8\times 50=40~N

Car 3⇒m=0.6, a=20 m/s²

\tt \sum F=0.6\times 20=12~N

Car 4⇒m=1, a=19~m/s²

\tt \sum=1\times 19=19~N

5 0
3 years ago
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