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3 years ago

In the following reaction, which element is oxidized and which is reduced?

Chemistry

1 answer:

bazaltina [42]3 years ago

3 0

Answer:

The answer to your question is: Iron oxidizes and Copper reduces

Explanation:

An element oxidizes when it loses electrons

An element reduces when it gains electrons

Then

Fe ⇒ Fe⁺² Now, is more positive, it loses electrons

Cu⁺² ⇒ Cu Now, is more negative, it gains electrons

Iron oxidizes and Copper reduces

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Which is longer, the carbon-oxygen single bond in a carboxylic acid or the carbon-oxygen bond in an alcohol?

Nikolay [14]

The carbon-oxygen single bond in a carboxylic acid

7 0

3 years ago

What is the mass of a sample of pure gold containing 3.00 x 1024 gold atoms?

miss Akunina [59]

Answer: The mass is 980.6g of Gold.

Explanation:

We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.

Using the Avogadro's number,

6.02 x 10^23 atoms of gold make up 1 mole of gold.

3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.

Now that we know the number of moles, we can then look for the mass using the formular:

Moles = mass/ molar mass

4.98 = mass / 196.9 (atomic mass of gold)

Making "mass" the subject of formula : mass = 4.98 x 196.9= 980.6g

8 0

3 years ago

What is the nuclear binding energy of an atom that has a mass defect of 5.0446

notsponge [240]

Answer:

Option C: 4.54 x $10^{11}$ KJ/mol of nuclei

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$ .

Explanation:

If mass defect is known, then nuclear binding energy can easily be calculated, here's how:

First step is to convert that mass defect into kg.

Mass defect = 5.0446 amu

Mass defect = 5.0466 x 1.6606 x $10^{-27}$

Because 1 amu = 1.6606 x $10^{-27}$ Kg.

Mass defect = 8.383 x $10^{-27}$ kg.

Now, we need to find out it's energy equivalent by using following equation:

Using the equation E = mc²:

where c= 3.00 x $10^{8}$ m/s²

E = (8.383 x $10^{-27}$ ) x (3.00 x $10^{8}$ )²

E = 7.54 x $10^{-10}$ J this energy is in Joules but nuclear binding energy is usually expressed in KJ/mol of nuclei. Let's convert it:

(7.54 x $10^{-10}$ Joule/nucleus)x(1 kJ/1000 Joule)x(6.022 x $10^{23}$ nuclei/mol) =

4.54 x $10^{11}$ kJ/mol of nuclei .

E = 4.54 x $10^{11}$ kJ/mol of nuclei . So, this is the nuclear binding energy of that atom, which is option C.

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$

4 0

3 years ago

Read 2 more answers

The most likely van't Hoff factor for an 0.01 m calcium iodide solution is

monitta

This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:

<h3>Van't Hoff factor:</h3>

In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.

In such a way, for calcium iodide, we write its ionization equation as shown below:

$CaI_2\rightarrow Ca^{2+}+2I^-$

Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.

Learn more about the van't Hoff factor: brainly.com/question/23764376

5 0

2 years ago

Help asap i have to get a good grade on this

baherus [9]

I think The answer is 34.5l

4 0

3 years ago