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25 days ago

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

1 answer:

7 0

Iodic acid undergoes partial dissociation to yield H+ and IO3- ions
. Let x represent the equilibrium concentration of H+, which is equivalent to the concentration of IO3- since they are produced in equal amounts. The concentration of HIO3 at equilibrium becomes 0.20 M - x
. The equation for dissociation is Ka = [H+] [IO3-] / [HIO3];
. Initially, [H+] is close to 0, [IO3-] is approximately 0, and [HIO3] stands at 0.20
. At the equilibrium state, [H+] and [IO3-] take the value of x, while [HIO3] is 0.20 - x
. This leads us to 0.17 = x² / (0.20 - x);
. Solving for x using the quadratic formula yields:
x = [H+] = 0.063 M and calculating pH gives us pH = - log [H+] = 1.2.

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lorasvet [2795]

Thanks for the responses ッ. (By the way, the answers are located at the bottom of the question if anyone hasn’t noticed them.

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16 days ago

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A mixture of KCl and KNO3

lorasvet [2795]

The percentage of KCl present in the mixture is approximately 40%If we consider a 100-gram sample of the mixture:ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equations:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From the first equation, we find x = 100 - y and substitute into the second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
Thus, x = m(KCl) = 100 g - 62.83 g = 37.17 g.

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29 days ago

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what is the net ionic equation with its physical states? (NH4)2CO3(aq)+Ca(ClO4)2(aq)⟶CaCO3(s)+2NH4ClO4(aq)

lions [2927]

Answer: The net ionic equation is $CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)$

Explanation:

A double displacement reaction involves the exchange of ions. Chemicals that dissolve in water are marked with the symbol (aq), while those that do not dissolve and remain solid are shown with (s) after their formulas.

$(NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)$

The ion-based representation of the equation is:

$2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)$

"Spectator ions" are the ions that do not participate in the chemical reaction, appearing on both sides of the equation in ionic form.

Ammonium and chlorate ions are present on both sides; thus, they do not factor into the net ionic equation.

Therefore, the net ionic equation is:

$CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)$

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1 month ago

You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation

Anarel [2989]

The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.

Thus, the rotation of the racemic mixture will be equal to 0°.

- This occurs because a racemic mixture contains equal proportions of both enantiomers.

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1 month ago

Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

KiRa [2933]

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

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1 month ago