Is liquid ammonia a household acid?

Which of the following forms a molecular solid? Which of the following forms a molecular solid? C, graphite C10H22 CaO gold

Damm [24]

Answer:

C10H22

Explanation:

Molecular Solids comprises of a Vander waal's force of attraction between the molecule. These forces are very weak when compared to ionic and covalent bond.

In Carbon, Carbon is not a molecule but an atom. One of it unique characteristics is that it forms bonds with other carbon atoms. This property is know as catenation. The bond between these carbon atoms is know as covalent bond.

Graphite is an allotrope of carbon. It exists as black , slippery, hexagonal crystals.The carbon atoms in graphite forms flat layers and are joined together by strong covalent bonds. Graphite can be used as lubricant in engines.

Gold (Au) is an element on the periodic table with atomic number 79 and a mass number 197. It exists as a metal. Most times Gold forms hydrogen bonds.

C10H22 is known as decane. It is the tenth compound formed in the series of alkane family( an organic unsaturated carbon chain family). Alkanes are aliphatic hydrocarbons. The forces of attraction between the alkane family are weak.In decane , their exists Vander waal's force which makes Decane C10H22 a Molecular Solid.

6 0

3 years ago

1. If 80.0 ml of 3.00 M HCl is used to make a 100. ml of dilute acid, what is the molarity

Aleonysh [2.5K]

We are given:

M1 = 3 Molar V1 = 80 mL

M2 = x Molar V2 = 100 mL

Finding the molarity:

We know that:

M₁V₁ = M₂V₂

where V can be in any units

(3)(80) = (x)(100)

x = 240/100 [dividing both sides by 100]

x = 2.4 Molar

3 0

2 years ago

In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl

STatiana [176]

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

5 0

3 years ago

How many grams of NO can be made from 50 g of NH3? 4 NH3 + 5 O2 → 6 H2O + 4 NO

Evgen [1.6K]

0.08 because 4mol NO divided by 50g gives you your answer

4 0

2 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

Is liquid ammonia a household acid?​

Is liquid ammonia a household acid?