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3 years ago

Please HURRYYYYYYYYYY

2 answers:

6 0

The periodic table is divided into 18 group. A group refers to a column of elements in the periodic table. All the elements in the same group share the same chemical and physical properties. Some of the groups are given specific names while the rest are identified by the first element in that group. The groups with specific names are as follows:
Group 1: Alkali metals
Group 2: Alkali earth metals
Group 11: Coinage metals
Group 15: Pnictogen
Group 16: Chalcogens
Group 17: Halogens
Group 18: Noble gases.

tia_tia [17]3 years ago

4 0

Elements of section s: are those elements whose valence shells are s¹ or s².

This section includes group 1 and 2.

Group 1: Alkalyne metals (Li, Na, K, Rb, Cs, Fr): react violently with water.

Group 2. Alkalyne earth metals (Be, Mg, Ca, Sr, Ba, Ra: react less violently with water.

Elements of both group 1 and group 2 shine, are conductive and are malleable and ductile.

Elements of section p: are those elements whose valence shells are p¹, or p², or p³, or p⁴, or p⁵, or p⁶.

This section includes the groups 13, 14, 15, 16, 17 and 18.

Most of those elements and non-metals except for B, Si, Ge, As, Sb, Te and Po which are metalloids (they have some characteristics of metals and some characteristics of non-metal).

Non metals are brittle and poor conductive.

Elements of section d is the biggest group. These are transition metals. In this group the electrons fill the type d orbitals.

This section includes groups 3 through 12.

The properties of these elements are those of metals: high conductivitiy, mallability and electrical conductivity.

Elements of section f are other transition metals (inner transition metals). They partially fill some f orbitals.

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levacccp [35]

Answer:

A pure substance has a definite chemical composition

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3 years ago

Vinegar produces carbon dioxide gas when combined with baking soda. Physical or chemical property?

Bas_tet [7]

Answer:

it's a chemical property

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3 years ago

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

Calculate moles of oxygen atoms in 0.68mol of KMnO4

Dominik [7]

Alright, so that means we have 0.68 mol of the compound

For each 1 mol of the compound, we have 4*1 oxygens (because there are four oxygens in the formula)
Therefore for each 0.68 mol of the compound, we have 4*0.68 moles of oxygen!

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3 years ago

Can you help me with acrostic poem?

kkurt [141]

An acrostic poem is a type of poetry where the first, last or other letters in a line spell out a particular word or phrase.

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3 years ago