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3 years ago

What is the electron configuration for Fe?

Chemistry

2 answers:

makkiz [27]3 years ago

7 0

The correct answer is C

blsea [12.9K]3 years ago

5 0

The answer is C) [Ar]4s2 3d6

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This is for a study guide for my semester final, and I have no clue how to answer this, please help, and please at least try to

user100 [1]

Positron emission = emission of a positron and a neutrino when a proton is convert into a neutron. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron

It's a spontaneous reaction for some nucleus.

eg:

Positron = e+

Neutrino=ve

O-15 --> N-15 + e+ +ve

Electron capture= A nucleus absorb an electron while a proton is convert in a neutron and emit a neutrino. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron

eg:

Al-26 +e- --> Mg-26 + ve

Electron capture and positron emission are two mechanisms to explain the decay of some unstable isotopes. Electron capture is usually observed when the energy difference between the initial and final state is low. Mainly because of the larger amount of kinetic energy need for the expulsion two particles with the positron emission mechanism.

7 0

3 years ago

Please help me the picture is above.

Andreas93 [3]

Answer:

Explanation:

bc inorganic compoud refers to all compound that do not contain carbons.

4 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$