Ask question

Ask question

All categories

3 years ago

5

What is the electron configuration of an electrically neutral alkaline earth metal? please explain

1 answer:

vlabodo [156]3 years ago

8 0

Answer:

The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns

You might be interested in

How many atoms of oxygen are in a molecule of glucose<br> (C6H12O6)

valentinak56 [21]

Answer:

It is 6

Explanation:

6 oxygen atoms

4 0

3 years ago

Read 2 more answers

Jill is doing a science experiment. After she mixes some baking soda into some

sertanlavr [38]

I think it's C. Please mark brainliest

8 0

3 years ago

A tree was chopped down and cut up in a local forest. The tree changed in shape.

Natasha2012 [34]

Answer would be C, Physical Change, it's similar to crumpling up a piece of paper

3 0

3 years ago

Read 2 more answers

How many unpaired electrons does calcium have? You may need to draw the orbital filling diagram for the last sublevel that conta

beks73 [17]

Answer:

0

Explanation:

Ca has 20 electrons

6 0

3 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Sladkaya [172]

Explanation:

The given balanced reaction is as follows.

$2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O$

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of $NH_{4}NO_{3}$ present will be as follows.

Moles of $NH_{4}NO_{3}$ = $\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}$

= $\frac{86000 g}{80.043 g/mol}$

= 1074.42 mol

Therefore, moles of $N_{2}$ , $O_{2}$ and $H_{2}O$ produced by 1074.42 mole of $NH_{4}NO_{3}$ will be as follows.

Moles of $O_{2}$ = $\frac{1}{2} \times 1074.42 mol$

= 537.21 mol

Moles of $N_{2}$ = $\frac{2}{2} \times 1074.42 mol$

= 1074.42 mol

Moles of $H_{2}O$ = $\frac{4}{2} \times 1074.42 mol$

= 2148.84 mol

Therefore, total number of moles will be as follows.

537.21 mol + 1074.42 mol + 2148.84 mol

= 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

PV = nRT

1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as $307^{o}C$ = 307 + 273 = 580 K)

V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.

3 0

3 years ago