Iron plus lead (||)sulfate react forming iron (||) sulfate plus lead

What is electron configuration of oxygen in its excited state

timofeeve [1]

Answer:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

OR

$2 : 6$

3 0

3 years ago

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Pre-Lab Study Questions / 9

jok3333 [9.3K]

Answer:

See detailed answer with explanation below.

Explanation:

Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.

Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.

Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.

In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.

The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.

4 0

3 years ago

How many Chorine atoms should bond with one Aluminum atom?

White raven [17]

Answer:

three chlorine atoms

Explanation:

8 0

2 years ago

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A sample of gas in a balloon has an initial temperature of 21 ∘C and a volume of 1310 L . If the temperature changes to 70 ∘C ,

bixtya [17]

Answer:

1528.3L

Explanation:

To solve this problem we should know this formula:

V₁ / T₁ = V₂ / T₂

We must convert the values of T° to Absolute T° (T° in K)

21°C + 273 = 294K

70°C + 273 = 343K

Now we can replace the data

1310L / 294K = V₂ / 343K

V₂ = (1310L / 294K) . 343K → 1528.3L

If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too

7 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago