What is the pH of a solution that contains 25 grams of hydrochloric acid (hcl) dissolved in 1.5 liters of water?

1 answer:

4 0

First we calculate the concentration of HCl:

Moles = mass / Mr
= 25 / 36.5
= 0.685 mol

Concentration = 0.685/1.5 = 0.457 mol / dm³

For a strong monoprotic acid, the concentration of hydrogen ions is equal to the acid concentration.

pH = -log[H+]
pH = -log(0.457)
= 0.34

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true or false an element's reactivity is determined by the number of protons found in an atom of the element

andreyandreev [35.5K]

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3 years ago

Please someone help me and all of the terms do not have to eh used

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Answer:

The correct answer is -

1. D. mold

2. A. Cast

3. F. trace fossil

4. C. original material fossils

5. B. fossil record.

Explanation:

Mold is the impression of an organism in a rock that occurs due to the complete dissolving of the organism and only an impression left in sedimentary rocks.

The cast is a fossil copy of an organism in a rock that takes place when water-containing minerals fill the molds and creates a three-dimension copy of the fossil.

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The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is appl

garik1379 [7]

Answer: Time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

Explanation:

The given data is as follows.

Surface area = 49.8 $cm^{2}$ ,

Density of gold = 19.3 $g/cm^{3}$ ,

Current = 3.15 A, thickness of gold layer = $1.2 \times 10^{-3} cm$

It is known that relation between volume, area and thickness is as follows.

V = Surface area × Thickness

= $49.8 \times 1.2 \times 10^{-3} cm$

= 0.05988 $cm^{3}$

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

$0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}$