All categories

3 years ago

What is the electron configuration of an electrically neutral atom of magnesium?

1 answer:

3 0

The electron configuration of an atom can be determined by identifying first the right order of the electron shells. The atomic number of Magnesium is 12 so we expect the electrons also sum up to 12. The condensed electron configuration is Ne 3s2. The complete configuration is 1s2 2s2 2p6 3s2.

You might be interested in

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

What is oxidized in a galvanic cell with aluminum and gold electrodes ?

k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0

3 years ago

8. Consider the element Scandium, atomic # 21.

yawa3891 [41]

Explanation:

Scandium has atomic number of 21. This means that in it's neutral state its going to have 21 electrons.

a) The full electronic configuration is given as;

1s2 2s2 2p6 3s2 3p6 3d1 4s2

The final electron is placed in the d orbital. The shell is 3d

(b) When scandium has a charge if +1, it has lost an electron. The total number of electrons would now be 21-1 = 20

The electronic configuration would be given as;

1s2 2s2 2p6 3s2 3p6 4s2

The electron in the 3d orbital would be removed.

4 0

3 years ago

Describe what happens when you take a antacid

lesya [120]

The antacid is alkaline so it neutralizes the excess acids in your stomach.

3 0

3 years ago

Compare the structure and function of rhizoids and stolons? please help I need this answer for science

ladessa [460]

structural difference between Rhizoids and stolons
Rhizoids are root like structure organs while stolons are a specialized type of horizontal above the ground

functional difference
Rhizoids absorb water and nutrient , They keep plant attached on the soil while stolons are propagation of the plant.

7 0

3 years ago