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White raven [17]
2 years ago
9

What property did mendeleev use to organize his periodic table?

Chemistry
1 answer:
Over [174]2 years ago
5 0
<h2>Answer:</h2>

Mendeleev grouped elements with similar chemical properties (especially valency) together and noted their mass as well. At first he only wrote down 9 elements:

Cl 35.5, K 39, Ca 40

Br 80, Rb 85, Sr 88

I 127, Cs 133, Ba 137

He added more elements to his first period table little by little, putting them into groups by valency and sorting them within their group by mass. He noted that the mass in most cases also increases by group within a period.

On 6 March 1869, he presented his work "The Dependence between the Properties of the Atomic Weights of the Elements" for the first time. It stated that:

The elements, if arranged according to their atomic weight, exhibit an apparent periodicity of properties.

Elements which are similar regarding their chemical properties either have similar atomic weights (e.g., Pt, Ir, Os) or have their atomic weights increasing regularly (e.g., K, Rb, Cs).

The arrangement of the elements in groups of elements in the order of their atomic weights corresponds to their so-called valencies, as well as, to some extent, to their distinctive chemical properties; as is apparent among other series in that of Li, Be, B, C, N, O, and F.

The elements which are the most widely diffused have small atomic weights.

The magnitude of the atomic weight determines the character of the element, just as the magnitude of the molecule determines the character of a compound body.

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Silver has a density of 10.5g/cm to the 3rd and gold has density of 9.3g/cm to the 3rd. Which would have a greater mass, 5cm to
Akimi4 [234]
The density of gold is 19.3 g/cm³

Ag= silver
Au= gold.
density (Ag)=10.5 g/cm³.
density (Au)=19.3 g/cm³

We have to calculate the mass of 5 cm³ silver:

10.5 g---------------------1 cm³
x---------------------------  5 cm³

x=(10.5 g  * 5 cm³) / 1cm³ ≈52.5 g

And we have to calculate the mass of 5 g gold.

19.3 g------------------------1 cm³
x-------------------------------5 cm³

x=(19.3 g * 5 cm³) / 1cm³≈96.5  g

Then 5 cm³ silver are 52.5 g of mass, and 5 cm³ gold are 96.5 g of mass, therefore,  gold would have a greater mass than silver.

answer:5 cm³ gold would have a greater mass than 5 cm³ silver.
3 0
4 years ago
How many milliliters of a 1M nitric acid solution are required to prepare 60mL of 6.7M solution?
Free_Kalibri [48]

Answer:

the number of milliliters of a 1M is 402mL

Explanation:

The computation of the number of milliliters could be determined by using the following formula

As we know that

V_1\times M_1 = V_2\times M_2

where,

V_1 and V_2 are the starting and final volumes

And, the M_1 and M_2 are the starting and the final molarities

Now the V_1 is

V_1 \times 1M = 60mL \times 6.7M

So, the V_1 is 402mL

Hence, the number of milliliters of a 1M is 402mL

4 0
3 years ago
When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver
slavikrds [6]

Answer:

m_{AgCl}=64.13gAgCl

Explanation:

Hello,

In this case, the undergone chemical reaction turns out:

2AgNO_3+MgCl_2-->2AgCl+Mg(NO_3)_2

In such a way, based on the reacting 21.3 g of magnesium chloride that are consumed the following stoichiometric procedure leads to the required grams of solver chloride precipitate consider their 1 to 2 mole relationship respectively:

m_{AgCl}=21.3gMgCl_2*\frac{1molMgCl_2}{95.211gMgCl_2}*\frac{2molAgCl}{1molMgCl_2}*\frac{143.32gAgCl}{1molAgCl}=64.13gAgCl

Best regards.

7 0
4 years ago
Determine the number of hydrogen atoms in an alkyne with one carbon–carbon triple bond and 4 carbon atoms.
rewona [7]

Answer:

There are 6 hydrogen atoms.

Explanation:

The formula for an alkyne that has only one triple bond is as follows:

  • CₙH₂₍ₙ₎₋₂

Where n is any natural number greater than 2.

For this problem, n=4. This would mean the formula is:

  • C₄H₂₍₄₎₋₂

Thus

  • C₄H₈₋₂
  • C₄H₆

Thus the answer is 6, there are 6 hydrogen atoms.

4 0
3 years ago
PLEASE HELP ASAP !!! WILL GIVE BRAINLIEST
galben [10]

Answer:

The atomic model started out as a large single ball, then with electrons attached to it. It then became  "planetary model" with electrons orbiting at specific levels, and then later one with a nucleus or multiple different particles like protons and neutrons, and electrons orbiting at general (but not specifically predictable) areas.

Explanation:

thank u :)

4 0
3 years ago
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