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kondor19780726 [428]
2 years ago
10

When we send electricity to pass through electrolysis jar we see gas bubble around both electrodes. Water electrolysis produces

46cm3 at anode.
A). What gas does it appear at each electrode?
B). What would you do to identify the gas in the test tubes at each electrode?
C). Calculate the volume of gas which appears at cathode.
Chemistry
1 answer:
Komok [63]2 years ago
5 0

A) Anode - hydrogen gas

Cathode-oxygen gas

B) Hydrogen gas - The gas burns with a pop sound

Oxygen gas - The gas rekindles a glowing splint

C) The volume of gas produced at the cathode is 23 cm^3

<h3>What is electrolysis?</h3>

Electrolysis is the process by which electric current is passed through a substance to effect a chemical change.

The electrolysis of water produces two volumes of hydrogen at the anode and one volume of oxygen at the cathode.

The hydrogen gas burns with a pop sound. It is easily identified by this test. Also, oxygen is known to rekindle a glowing splint and this is a commonly used test for oxygen gas.

Since the ratio of hydrogen produced at the anode to oxygen produced at the cathode is 2:1 by volume; then when 23 cm^3 of oxygen is produced at the cathode, 46 cm^3of hydrogen is produced at the anode.

Learn more about the electrolysis here:

brainly.com/question/12054569

#SPJ1

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4. Consider the following statement made in Chapter 4 of the text: "... consider that if the
AveGali [126]

The claim: "If the nucleus were the size of a grape, the electrons would be one mile away on average" is reasonably accurate because the ratios between the nucleus's sizes and the distances (between electrons and nucleus) for the two given examples are in the same order of magnitude.      

To know if the claim is accurate we need to calculate the ratio of the size of the nucleus (the same as a grape) and the distance between the electrons and the nucleus for example 1 (r₁):  

r_{1} = \frac{s_{1}}{d_{1}}    (1)

and to compare it with the ratio of the size and the distance given in example 2 (r₂):

r_{2} = \frac{s_{2}}{d_{2}}    (2)

<em>Where:</em>

s₁: is the size of the nucleus (like the size of a grape)

d₁: is the distance between electrons and nucleus of example 1 = 1 mile

s₂: is the average diameter of the nucleus  = 10⁻¹³ cm

d₂: is the average distance between electrons and nucleus of example 2 = 10⁻⁸ cm

Assuming that the diameter of a grape is 3 cm (in a spherical way), the ratio of the <u>first example</u> is (eq 1):

r_{1} = \frac{3 cm}{1 mi*\frac{160934 cm}{1 mi}} = 1.86 \cdot 10^{-5}

Now, the ratio of the <u>second example</u> is (eq 2):

r_{2} = \frac{10^{-13} cm}{10^{-8} cm} = 1.00 \cdot 10^{-5}              

Since r₁ and r₂ are in the same order of magnitude (10⁻⁵), we can conclude that the given claim is reasonably accurate.      

You can learn more about the nucleus of an atom here: brainly.com/question/10658589?referrer=searchResults

I hope it helps you!                

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3 years ago
Which element easily loses one electron to form a positive ion? (AKS 1c/DOK 1)
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Answer:

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Explanation:

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8 0
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Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if 2 moles of pentane (C5H12) were supplied, and an unlimited amount
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We have the following combustion of pentane (C₅H₁₂):

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Knowing the chemical reaction we devise the following reasoning:

if         1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O

then    2 moles of pentane C₅H₁₂ produces X moles of water H₂O

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Learn more about:

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brainly.com/question/7295137

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#learnwithBrainly

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