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4 years ago

The vertical columns on the periodic table are called groups, how many groups are there on the periodic table?

1 answer:

5 0

8 groups, 1 to 7 and most people call group 8 group 0 because the noble gases in group 0 have the outer electron shell full, so for example argon has 18 electrons, so it’s structure is 2.8.8 with its outer shell full. Also the number of the last shell defies what group it’s in, for example sodium is in group 1 so it’s structure is 2.8.1. Hope this helps.

You might be interested in

Why is it difficult to accurately represent ammonia and methane on paper?

andrew-mc [135]

Answer:

The correct answer is because the molecular structure.

Explanation:

The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.

In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.

Have a nice day!

6 0

3 years ago

Another name for a "oxidation-reduction" reaction is

anzhelika [568]

Redox (red for reduction and ox for oxidation)

8 0

4 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin

Lerok [7]

<u>Answer:</u> The percentage yield of HF is 73.36 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %

4 0

3 years ago

Determine which of the following statements is true:

alukav5142 [94]

Answer:

A. When an anhydrous salt becomes hydrated, it can change color.

Explanation:

The cobalt (II) chloride (CoCl₂) have a blue color in the anhydrous form while after hydration, most commonly it will form the hexahydrate, it will change its color to violet.

B. not really, if we take the cobalt (II) chloride in the anhydrous form the bond between cobalt and the chloride it is ionic.

C. nope, the mass is changed depending on the lost water mass

D. no, the mass will increase and sometimes as in the case of cobalt (II) chloride the color will change, so you have changes of the physical properties of the substance.

3 0

3 years ago