A element that starts with the letter c (from the periodic table)

What is the electron configuration for vanadium (V)? The Periodic Table A. 1s22s22p63s23p64s24d3 B. 1s22s22p63s23p63d5 C. 1s22s2

kramer

The electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 (option D).

<h3>What is electronic configuration?</h3>

Electronic configuration is the the arrangement of electrons in an atom, molecule, or other physical structure like a crystal.

Vanadium is the 23rd element on the periodic table and has chemical symbol V with atomic number 23. It is a transition metal, used in the production of special steels.

This suggests that the electronic configuration of Vanadium will be written as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Therefore, the electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3.

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5 0

2 years ago

What is the oxidation number of chlorine in Al(ClO4)3? Please give me unique work! THANKSS

sveta [45]

It would be +7 oxidation number.
[All compounds are zero]
Al is in group III [ or 13 depending on the system ] so the oxidation state is 3.
You have 12 O at -2 each for-24. That leaves you at -24+3=-21. So, 3 Cl atoms must be +21 to balance the -21 there which makes +21/3 or +7 for each Cl.

3 0

3 years ago

In the Bohr model of the atom, the electrons move in fixed, ________ paths around a dense, positively-charged nucleus. On the ot

trapecia [35]

Answer:

Answers are in parentheses.

In the Bohr model of the atom, the electrons move in fixed, (circular) paths around a dense positively-charged nucleus. On the other hand, the quantum mechanical model shows the probability of finding an electron as a (cloud) of negative charge.

6 0

3 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ 63 g

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

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7 0

3 years ago

Ammonia (NH3) is one of the most common chemicals produced in the united states. it is used to make fertilizer and other product

Greeley [361]

Answer:

a) N2 is the limiting reactant

b) 1215.9 grams NH3

c) 2283.6 grams H2

Explanation:

Step 1: Data given

Mass of N2 = 1000 grams

Molar mass N2 = 28 g/mol

Mass of H2 = 2500 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) +3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 1000 grams / 28.0 g/mol

Moles N2 = 35.7 moles

Step 4: Calculate moles H2

Moles H2 = 2500 grams / 2.02 g/mol

Moles H2 = 1237.6 moles

Step 5: Calculate limiting reactant

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

N2 is the limiting reactant. There will be consumed 35.7 moles.

H2 is in excess. There will react 3*35.7 = 107.1 moles

There will remain 1237.6 - 107.1= 1130.5 moles H2

This is 1130.5 * 2.2 = 2283.6 grams H2

Step 6: Calculate moles NH3

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

For 35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 71.4 moles * 17.03 g/mol

Mass NH3 = 1215.9 grams NH3

7 0

3 years ago