All categories

3 years ago

Identify the group number of following oxides/hydrides in Mendeleev

Chemistry

1 answer:

Ivan3 years ago

5 0

Answer:

See explanation

Explanation:

The oxides or hydrides are formed by exchange of valency between the two atoms involved. The group of the atom bonded to oxygen or hydrogen in the binary compound can be deduced by considering the subscript attached to the oxygen or hydrogen atom.

Now let us take the journey;

R2O3- refers to an oxide of a group 13 element, eg Al2O3

R2O - refers to an oxide of group a group 1 element e.gNa2O

RO2 - refers to an oxide of a group 14, 15 or 16 element such as CO2, NO2 or SO2

RH2 - refers to the hydride of a group 12 element Eg CaH2

R2O7 - refers to an oxide of a group 17 element E.g Cl2O7

RH3- refers to a hydride of a group 13 element E.g AlH3

You might be interested in

Alpha particles are helium nuclei. electromagnetic waves. electrons. neutrons.

Firdavs [7]

Alpha particles are helium nucleus ! so answer is A !

it is made by ionization of helium, when both the electrons are out , then two unit of positive charge is one helium and thus called alpha particle !

4 0

3 years ago

Define Titanium and what it is used for

Kaylis [27]

Chemical element of atomic number 22, a hard silver-gray metal of the transition series, used in strong, light, corrosion-resistant alloys.These alloys are mainly used in aircraft, spacecraft and missiles because of their low density and ability to withstand extremes of temperature.

8 0

3 years ago

Aluminum is more reactive than iron, yet it is used today for a variety of applications in which iron would corrode (cans, rain

marin [14]

Aluminum is more reactive than iron

but it forms Al-oxide, that form a thin layer on the surface of Al, and protect Al from reaction with water.

Answer: Very unreactive aluminum oxide forms a thin layer on aluminum.

6 0

3 years ago

Free fire loveer...................

rosijanka [135]

Answer:

Thanks! :) Have a great day!

Explanation:

5 0

3 years ago

Androstenedione, which contains only carbon, hydrogen, and oxygen, is a steroid hormone produced in the adrenal glands and the g

Gennadij [26K]

Answer: The molecular formula for androstenedione is, $C_{19}H_{26}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=5.527g$

Mass of $H_2O=1.548g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 5.527 g of carbon dioxide, $\frac{12}{44}\times 5.527=1.507g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.548 g of water, $\frac{2}{18}\times 1.548=0.172g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.893g)-[(1.507g)+(0.172g)]=0.214g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.0133$ moles.

For Carbon = $\frac{0.126}{0.0133}=9.5$

For Hydrogen = $\frac{0.172}{0.0133}=12.9\approx 13$

For Oxygen = $\frac{0.0133}{0.0133}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is $C_{19}H_{26}O_2$

The empirical formula weight of $C_{19}H_{26}O_2$ = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{286.4}{286}=1$

Molecular formula = $C_{19}H_{26}O_2$

Therefore, the molecular formula for androstenedione is, $C_{19}H_{26}O_2$

3 0

3 years ago