I need help with number 5 part b please!

Which statement is always true of the cathode in an electrochemical cell? Reduction occurs here. It is considered the "negative"

lara31 [8.8K]

Answer:

It Is Considered The "negative" Electrode

Explanation:

An electrochemical cell is an electrolytic cell that drives a non-spontaneous redox reaction through the application of electrical energy. This cell is used to decompose chemical compounds, in a process called electrolysis. An electrode at which reduction take place is called the cathode. In reduction, electrons travel toward the site of reduction such that the negative charge is on the cathode.

7 0

3 years ago

Fluorine gas reacts with aqueous iron (II) iodine to produce iron (II) fluoride and iodine liquid What is the balanced chemical

Andru [333]

Answer:

F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)

Explanation:

Our reactants are:

F₂ → Fluorine gas, a dyatomic molecule

FeI₂ → Iron (II) iodine

Our products are:

I₂ → Iodine

FeF₂ → Iron (II) fluoride

Then, the reaction is:

F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)

We see it is completely balanced.

6 0

3 years ago

How many objects are in a mole of objects?

lakkis [162]

Your Question: {How many objects are in a mole?}

Helpful Knowledge: (We Know the amount in an object: 12g or C^12)

{A number of objects that are in a mole of objects?}

Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³

Which 6.02 × 10²³ is an Avogadro's Number.

So it depends on how many objects you have.

So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.

Anywho, your answer would be 6.02 x 10²³ x n.
N would equal the number of objects you're calculating.

Final Answer: 6.02 x 10²³ x (n) = (Your Answer)

Hope this helps! Have a great day. If you need anything else, feel free to hope right in my inbox. Or comment below. ↓

6 0

3 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0

3 years ago

If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?

bija089 [108]

<h3>Molar mass:-</h3>

$\\ \sf\longmapsto HC_2H_3O_2$