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pychu [463]
3 years ago
6

Match the following parts of an atom to their description.

Chemistry
1 answer:
Burka [1]3 years ago
5 0

Answer:

Neutron - non charged particle in nucleus

nucleus - composed of proton and Neutron

electron - negatively charged particle that moves around the nucleus

proton - positively charged particle in the nucleus.

Explanation:

these the are definitions of the terms

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Does anyone know how to write out the formula for these?? not just the answer x
Tju [1.3M]
Idk how to write the formula I had this same question
5 0
3 years ago
The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and
Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>

<em />

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

<em>Total moles Ag⁺:</em>

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

<em>Moles of Ag⁺ in the precipitate:</em>

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

<em>Moles AsO₄ = Moles As:</em>

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

<em>Moles As₂O₃:</em>

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

<em>Mass As₂O₃:</em>

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

<em />

7 0
3 years ago
An atom of Fe has 30 neutrons. What is the mass number of this atom?
SVETLANKA909090 [29]

Answer:

56

Explanation:

The mass number of an atom is a sum of its number of protons and neutrons. Iron (Fe) has an atomic number of 26. This means that it has 26 protons. Since we know it has 26 protons and are told it has 30 neutrons. The mass number of this atom of Iron is 26+30 = 56

Hope this helped!

7 0
3 years ago
Which description is true of 0.02 M Ca(OH)2 if Ca(OH)2 completely dissociates in water? A. It’s a dilute strong acid. B. It’s a
Sphinxa [80]

Answer: B. It’s a dilute strong base.


Explanation:


1) Definition of acids and bases: as per Bronsted-Lowry model, an acid is a substance that donates hydrogen ions and a base is a substance that accepts hydrogen ions.

Ca(OH)₂ does not have hydrogen ions to donate, but it can accept hydrogen ions to form H₂O according to this equation: H⁺ + OH⁻ → H₂O.

Hence,  Ca(OH)₂ is a base.


2) Definition of strong base: a strong base is a base that dissociates completely into metallic and hydroxide ions in aqueous solutions, while a weak base dissociates partially.


Hence, Ca(OH)₂ is a strong base.


3) Definition of dilute: it refers to a solution meaning that the substance is not pure and the concentration is low. Since, the solution the Ca(OH)₂ is 0.02 M means that it is dilute.


Therefore, we have found that the description of 0.02 M Ca(OH)₂ is that is is a dilute strong base (option B).


5 0
3 years ago
The reaction H2(g)+F2(g)?2HF(g) is spontaneous at all temperatures. A classmate calculates the entropy change for this reaction
Pachacha [2.7K]

Answer:

The correct answer is there is a mistake in the calculation. The second law of thermodynamics state that in any spontaneous process there is an increase in the entropy of the universe.

Explanation:

According to the second law a reaction will occur in a system spontaneously if the total entropy of both system and surrounding increases during the reaction.That means in case of spontaneous reaction entropy change is always positive.

     But according to the question the reaction H2+F2=2HF is spontaneous in all temperature.So according to the second law of thermodynamics i can say that my classmate made a mistake in calculation that"s why his result for entropy change comes negative.

6 0
3 years ago
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