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3 years ago

Which electron configuration represents a violation of the pauli exclusion principle?

2 answers:

7 0

The electron configuration that represents a violation of the pauli exclusion principle is:

1s: ↑↓
2s: ↑↑
2p: ↑
The Pauli exclusion principle refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.

Flauer [41]3 years ago

5 0

Answer:

Explanation:

The electron configuration that represents a violation of the pauli exclusion principle is:

1s: ↑↓

2s: ↑↑

2p: ↑

The Pauli exclusion principle refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.

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The symbol for the element in period 2, group 13

denis-greek [22]

Element: Boron
Symbol: B
So answer is B

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2 years ago

An element has three naturally occurring isotopes. Use the information below to calculate the weighted average atomic mass of th

Mariulka [41]

It is isotope ok!!!!!

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3 years ago

A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132

Yakvenalex [24]

First - you need the empirical formula.

So, assume you have 100 g of the compound.

If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.

54.53 g C (1 mole C / 12.01 g C) = 4.540

9.15 g H (1 mole H / 1.008 g H) = 9.077

36.32 g O (1 mole O / 15.9994 g O) = 2.270

Take the smallest number found and divide the others by it to get the empirical formula.

4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.

So, that gives you the empirical formula of C2H4O.

Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.

132/44 = 3.

So, 3 (C2 H4 O) = C6H12O3 = molecular formula.

6 0

3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1