What is a small description of a molecule

A 5 g sample of pure gold has a density of 19.3 g/ml. Your friend purchased a gold ring that was made of 25 g of pure gold. What

patriot [66]

Answer:

96.5 g/ml

Explanation:

If 5g is 19.3 then 25g is 19.3x5 which is 96.5 g/ml

7 0

3 years ago

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How many hydrogen (H) atoms are in 1 molecule of (NH4)2S ?

Alex73 [517]

Im not sure but it could be 4 (H)

3 0

3 years ago

What is the systematic name for TiCl2

Anarel [89]

Answer:

66228

Molecular Formula: TiCl2 or Cl2Ti

Chemical Names: Titanium chloride (TiCl2) 10049-06-6 TiCl2 Titanium(II) chloride dichlorotitanium More...

Molecular Weight: 118.77 g/mol

Dates: Modify: 2019-08-10 Create: 2005-03-26

Explanation:

66228

Molecular Formula: TiCl2 or Cl2Ti

Chemical Names: Titanium chloride (TiCl2) 10049-06-6 TiCl2 Titanium(II) chloride dichlorotitanium More...

Molecular Weight: 118.77 g/mol

Dates: Modify: 2019-08-10 Create: 2005-03-26

8 0

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:

a)

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.

b)

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.

c)

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. This means that citric acid will not react everything, leaving an excess.

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

Which of the following processes will determine the number of moles in a sample? Dividing the mass of the sample by Avogadro's n

Olenka [21]

The answer is: Dividing the number of molecules in the sample by Avogadro's number.

The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).

Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).

The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.

For example:

N(Ba) = 2.62·10²³; number of atoms of barium.

n(Ba) = N(Ba) ÷ Na.

n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.

n(Ba) = 2.158 mol; amount of substance of barium.

6 0

3 years ago