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3 years ago

Could someone explain what oxidation numbers are and what these questions are asking of me?

Chemistry

1 answer:

Oxana [17]3 years ago

6 0

The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation of an atom in a chemical compound.

Explanation:

The oxidation number of an atom is the charge that atom would have if the compound was composed of ions. 1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. The oxidation number of simple ions is equal to the charge on the ion.

The oxidation number of a mono atomic ion equals the charge of the ion. The oxidation number of H is +1, but it is -1 in when combined with less electro negative elements. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides. The oxidation number of a Group 1 element in a compound is +1.

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Answer:

Scientific models are used to explain and predict the behaviour of real objects or systems and are used in a variety of scientific disciplines, ranging from physics and chemistry to ecology and the Earth sciences.

Explanation:

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3 years ago

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The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago

Half of the first 18 elements have an odd number of electrons, and half have an even number. Show why atoms of these elements ar

asambeis [7]

Answer:

Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.

Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.

Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.

Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.

The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.

Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.

Explanation:

3 0

3 years ago

The below given compound is slightly soluble in water.

nordsb [41]

Answer:

an alkali

Explanation:

an alkali is a base that is slightly soluble in water

4 0

3 years ago