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2 years ago

Where are the most active nonmetals located on the periodic table

Chemistry

2 answers:

natulia [17]2 years ago

5 0

Answer:

Group 7

Explanation:

Elements in the group 7 of the periodic table are known as the halogens. They are the most active/reactive nonmetals because they have just one electron to receive in order to achieve there octet configuration. They react easily with elements in group 1 (in the same proportion) since group 1 elements have just one electron in there outermost shell. Halogens easily form acids with hydrogen by sharing the only electron hydrogen has with its own valence electron via covalent bonding. They however undergo ionic bonding with the remaining members of group 1. They also react with metals in group 2, 3 and 4 in different proportion.

AURORKA [14]2 years ago

4 0

Answer:

The most active nonmetals belong to the halogen family, which sits to the left of the noble gases on the right side of the periodic table. The halogens are so reactive that they are never found in nature by themselves. The elements fluorine, chlorine, bromine, iodine and astatine make up the halogen family.

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A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$ where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$

7 0

2 years ago

An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

ozzi

In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$ .

4 0

3 years ago

A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? Use the peri

meriva

The concentration of NaOH is 25.00% by mass, it means that 25.00% of the mass of the solution is of NaOH. Hence:

$m_{NaOH}=25.00\%\times m_{solution}\Longrightarrow m_{NaOH}=\dfrac{25}{100}\times 0.250~g\iff\\\\\boxed{m_{NaOH}=0.0625~g}$

7 0

3 years ago

Read 2 more answers

As a sodium atom is oxidized the number of protons in its nucleus

lara31 [8.8K]

Answer:

The proton remains the same.

Explanation:

Oxidation is simply defined as the loss of electron(s) during a chemical reaction either by an atom, molecule or ion.

Oxidation is strictly on the transfer of electron(s) and not proton.

A metal that undergoes oxidation still has its protons intact otherwise it will not be called the ion of the metal since atomic number is called the proton number.

Sodium (Na) undergoes oxidation as follow:

Na —> Na+ + e-

Na is called sodium metal.

Na+ is called sodium ion.

Na has 11 electrons and 11 protons

Na+ has 10 electrons and 11 protons

From the above illustration, we can see that the protons of Na and Na+ are the same why their electrons differ because Na+ indicates that 1 electron has been loss or transferred.

6 0

3 years ago

Which two properties are explained by the pool-of-shared-electrons model for metals?

Svetllana [295]

I believe it is C but if not that A

5 0

3 years ago