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anzhelika [568]

3 years ago

10

Which trend is observed as the first four elements in group 17 on the periodic table are considered in order of increasing atomi

c number

1 answer:

s344n2d4d5 [400]3 years ago

6 0

ANSWER:

The melting and boiling points increase in order of increasing atomic number.

The size of the nucleus increases in order of increasing atomic number.

Ionization energy decreases in order of increasing atomic number.

Electronegativity decreases in order of increasing atomic number.

Electron Affinity decreases in order of increasing atomic number.

The reactivities decrease in order of increasing atomic number.

EXPLANATION:

NAME MELTING POINT BOILING POINT

Fluorine -220 -188

Chlorine -101 -35

Bromine -7.2 58.8

Iodine 114 184

Melting and Boiling points increase as shown above.

NAME COVALENT RADIUS IONIC RADIUS

Fluorine 71 133

Chlorine 99 181

Bromine 114 196

Iodine 133 220

Size increases as shown above.

NAME FIRST IONIZATION ENERGY

Fluorine 1681

Chlorine 1251

Bromine 1140

Iodine 1008

Ionization energy decreases as shown above.

NAME ELECTRONEGATIVITY

Fluorine 4

Chlorine 3

Bromine 2.8

Iodine 2.5

Electronegativity decreases as shown above.

NAME ELECTRON AFFINITY

Fluorine -328.0

Chlorine -349.0

Bromine -324.6

Iodine -295.2

Electron affinity decreases as shown above.

REACTIVITY

The reactivities of the halogens decrease. This is due to the fact that atomic radius increases in proportion with an increase of electronic energy levels. This decreases the pull for valence electrons of other atoms, minimizing reactivity.

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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

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

Formula used :

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

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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