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Lesechka [4]
3 years ago
7

Please help!!Show the electronic configuration of magnesium atom​.

Chemistry
2 answers:
ch4aika [34]3 years ago
6 0

Answer:

Neon

[He]2s²2p⁶

Explanation:

just search that and u will see it -_-

Galina-37 [17]3 years ago
3 0

Explanation:

Magnesium has atomic number 12. It will be distributed in K, L, M shell in the following way:

K shell can accommodate a maximum of 2 electrons.

L shell can accommodate a maximum of 8 electrons and

M shell will accommodate 2 electrons

So the configuration becomes 2, 8,3

You might be interested in
Density of 2.7 g/ml and volume of 35.6 ml what is the mass
Anettt [7]
The math is set up like

35.6 ml * 2.7 g/ 1 ml

which will leave you with

96.12 g
3 0
4 years ago
Calculate the pH of a 2.3 10-3M[OH'] solution.
disa [49]

Answer:

11.33

Explanation:

-log(2.3x10^-3) = 2.67

14-2.67

- Hope this helped! Let me know if you need a further explanation.

5 0
4 years ago
What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?
aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0
3 years ago
For each element, predict where the "jump " occurs for successive ionization energies. (For example, does the jump occur between
vichka [17]

Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron

The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

The electronic configuration of Lithium is 1s2 2s1

There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.

8 0
4 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
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