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KatRina [158]
3 years ago
6

Write a paragraph (5-7 sentences) describing pH and conductivity.

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
4 0

Answer:

pH is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution. The pH scale usually ranges from 0 to 14. Aqueous solutions at 25°C with a pH less than 7 are acidic, while those with a pH greater than 7 are basic or alkaline. A pH level of 7.0 at 25°C is defined as "neutral" because the concentration of H3O+ equals the concentration of OH− in pure water. On the other hand, electrical conductivity is a non-specific measurement of the concentration of both positively and negatively charged ions within a sample. So the short answer to the question is as follows, the presence of any hydrogen ions present in a substance will impact the pH level and most probably influence conductivity levels. However, hydrogen ions make up only a small part of the ion concentration measured by a conductivity meter.

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What happens when you put metal into hot galium?
Mice21 [21]

Answer:

Aluminum Gallium Reaction- Gallium reacts with Aluminum to create an Aluminum alloy that crumbles at the touch. Gallium attacks many metals including Aluminum & Steel by diffusing into the grain boundaries making them extremely brittle. Gallium easily alloys with many metals in low quantities

Explanation:

3 0
3 years ago
Please answer
lutik1710 [3]

Answer:

See Explanation

Explanation:

8.  3, 1, -1, +1/2 => 3pₓ¹ => Aluminum

9.  4, 2, +1, +1/2 => 4d₁¹ => Chromium

10. 6, 1, 0, -1/2 => 6p₀² => Argon

11.  4, 3, +3, -1/2 => 4f₊₃² => Lutetium

12. 2, 1, +1, -1/2  => 2p₊₁² => Neon

8 0
3 years ago
A) If Kb for NX3 is 4.5×10^−6, what is the pOH of a 0.175 M aqueous solution of NX3 ?
sukhopar [10]

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

<h3>What is pH?</h3>

pH is a measure of how acidic/basic water is.

A)

NX_3 + H_2O →NHX_3^+ + OH^-

Kb = 4.5 x10^-6

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

x² = Kb x concentration of (NH₃)

x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

Learn more about pH here:

brainly.com/question/12353627

#SPJ1

5 0
2 years ago
Differentiate between a precipitate and an aqueous solution
antiseptic1488 [7]

Answer:

A precipate is a solid while an aqueous solution is liquid.

Explanation:

A precipitate is a solid which separates after a chemical reaction occurs. It is the solid product of the reaction.

An aqueous solution is formed when a substance is dissolved in water.

6 0
4 years ago
1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l)
Zanzabum

Answer:

The answer to your question is: yield = 56.27%

Explanation:

Data

             CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br

                18.54 ml 1-butanol            15.65 g of 1-bromobutane

% yield = ?

density = 0.81 g/ml

MM = 74 g  1- butanol

MM = 137 g 1-bromobutane

Process

Calculate mass of 1- butanol

              density = mass/volume

              mass = density x volume

              mass = 0.81 x 18.54

              mass = 15.02 g of 1-butanol

Theoretical yield

              74 g of 1- butanol -----------------  137 g of 1-bromobutane

              15.02 g of 1- butanol -------------   x

              x = (15.02 x 137) / 74

              x = 27.81 g of 1-bromobutane

% yield = experimental yield / theoretical yield x 100

% yield = 15.65 / 27.81 x 100

% yield = 56.28

6 0
4 years ago
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