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White raven [17]
3 years ago
10

!!! PLEASE HURRY I HAVE 15 MINUTES Describe the proccess which Mendeleev developed the 1 Periodic Table of the Elements

Chemistry
1 answer:
andrezito [222]3 years ago
4 0

In 1869, just five years after John Newlands put forward his law of octaves, a Russian chemist called Dmitri Mendeleev published a periodic table. Mendeleev also arranged the elements known at the time in order of relative atomic mass, but he did some other things that made his table much more successful.  

i hope this helps in some way!

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What is the answer for those questions plz ,help
marysya [2.9K]

Answer:

1) 1.51 × 10²³  particles of Mg

2) 0.54 × 10⁻³  moles

Explanation:

Given data:

1)

Number of moles of Mg = 0.250 mol

Number of representative particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  particles

0.250 mol × 6.022 × 10²³  particles / 1 mol

1.51 × 10²³  particles of Mg

2)

Given data:

Number of moles of lead = ?

Number of atoms of lead = 3.25×10²⁰ atoms

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  atoms

1 mol × 3.25×10²⁰ atoms/ 6.022 × 10²³  atoms

0.54 × 10⁻³  moles

6 0
3 years ago
The activation energy for the gas phase isomerization of cyclopropane is 272 kJ. (CH2)3CH3CH=CH2 The rate constant at 718 K is 2
MrMuchimi
The Arrhenius equation relates activation energy to reaction rates and temperature:
ln (k2 / k1) = (E / R) * (1/T1 - 1/T2), where E is activation energy of 272 kJ, R is the ideal gas constant (we use the units of 0.0083145 kJ/mol-K for consistency, to cancel out the kJ unit), we let T1 = 718 K and k1 = 2.30 x 10^-5, and T2 = 753 K and k2 be the unknown.
ln (k2 / 2.30x10^-5) = (272 kJ / 0.0083145 kJ/mol-K) * (1/718 - 1/753)
k2 = 1.91 x 10^-4 /s

5 0
3 years ago
A sample of gas at 15 atmospheres and 445 K is cooled to 250 K. What pressure will the gas be under after it is cooled? Round to
faltersainse [42]

Answer:

Explanation: 8.43

7 0
3 years ago
How many moles of nitrogen are in 2.31L of N2
PtichkaEL [24]

Hey there!

N₂ has a density of 1.25 grams per liter.

We have 2.31 liters of N₂.

2.31 x 1.25 = 2.8875

We have 2.8875 grams of N₂.

Find molar mass of N₂:

N: 2 x 14

------------

         28 g/mol

2.8875 ÷ 28 = 0.10

There is 0.10 mol of nitrogen in 2.31 L of N₂.

Hope this helps!

3 0
3 years ago
Who are you voting for? if u r and if u could
stiks02 [169]

Answer:

Biden

Explanation:

4 0
3 years ago
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