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liq [111]
3 years ago
7

O que é um número quântico?

Chemistry
1 answer:
Lera25 [3.4K]3 years ago
7 0
I'm going to put it in chat.Hope this help.
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For some transformation having kinetics that obey the Avrami equation (Equation 11.17), the parameter n is known to have a value
Alex787 [66]

Answer:

t ≈ 235.47 secs

Explanation:

<u>Calculate how long it will take the transformation to go to 88% completion</u>

Given that Avrami equation ( <em>y </em>)= 1 - exp( - kt^n )

n = 2.3

t = 145 secs

reaction = 50%  after t = 145 secs

Reaction = 88%  after t = ?

attached below is a detailed solution

3 0
3 years ago
List five examples of matter and five examples that are not matter. Explain your answer.
ASHA 777 [7]

Answer:

An apple,a person.a table,a computer,wood because they occupies space

non matter=Light, heat, kinetic and potential energy, and sound are non-matter because they are massless

3 0
3 years ago
Describe the formation of a sodium ion.
Dvinal [7]
Sodium ion- it has 1 valence electron so its easy for it to loose this electron. Thus, it forms sodium ion by loosing its valence elctron while reacting.It forms Na+. ... Oxide ion- it has 6 valence electrons so it gains 2 electrons lost by a metal atom and reacts with it.
5 0
3 years ago
30 ml of. 150 m cacl2 is added to a 15 ml of. 100 m agno3
ycow [4]

The mass of the formed precipitate of AgCl in the reaction is 1.29 grams.

<h3>How do we find moles from molarity?</h3>

Moles (n) of any substance from molarity (M) will be calculated by using the below equation:
M = n/V, where

V = volume in L

Given chemical reaction is:
2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

  • Moles of CaCl₂ = 0.150M × 0.03L = 0.0045 moles
  • Moles of AgNO₃ = 0.100M × 0.015L = 0.0015 moles

From the stoichiometry of the reaction, mole ratio of AgNO₃ to CaCl₂ is 2:1.

0.0015 moles of AgNO₃ = reacts with 1/2×0.0045 = 0.00075 moles of CaCl₂

Here CaCl₂ is the limiting reagent, and formation of precipitate depends on this only.

Again from the stoichiometry of the reaction:

0.0045 moles of CaCl₂ = produces 2(0.0045) = 0.009 moles of AgCl

Mass of 0.009 moles AgCl will be calculated as:

n = W/M, where

  • W = required mass
  • M = molar mass = 143.45 g/mol

W = (0.009)(143.45) = 1.29g

Hence required mass of precipitate is 1.29 grams.

To know more about moles & molarity, visit the below link:
brainly.com/question/24322641

#SPJ4

8 0
2 years ago
What 3 athletic events do not use the SI unit of measurement
bazaltina [42]

Answer:

Long jump doesn't use the SI unit of measurement

3 0
3 years ago
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