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3 years ago

Estimate the optimal number of neutrons for a nucleus containing 70 protons.

Chemistry

2 answers:

denpristay [2]3 years ago

7 0

The optimal number of neutrons for a nucleus containing 70 protons is 130

<h3>Further explanation </h3>

In chemistry, the nucleus is the positively charged center of the atom that consist of protons and neutrons. It's also known as the "atomic nucleus"

The neutron is a subatomic particle with no net electric charge and a mass slightly greater than that of a proton. The neutron is have symbol n or n⁰. Protons and neutrons constitute the nuclei of atoms.

Neutron is a type of atomic particle which does not possess any kind of charge on it. It has neutral behavior toward atoms since it does not possess any kind of charges but it contributes to the mass of the atoms.

Atomic number = 70

Atomic weight = 173

Atomic weight = number of protons + number of neutrons

Atomic number = number of protons

Therefore, the number of protons = 70

Number of neutrons = atomic weight - number of protons

Number of neutrons = 173 - 70 = 103

Optimal number of neutrons = 130

<h3>Learn more</h3>

Learn more about neutrons brainly.com/question/1823582
Learn more about nucleus brainly.com/question/2345919
Learn more about protons brainly.com/question/2166462

<h3>Answer details</h3>

Grade: 9

Subject: chemistry

Chapter: the optimal number of neutrons

Keywords: protons, nucleus, neutrons

DochEvi [55]3 years ago

6 0

Atomic number 70

weight 173

number of neutron = weight - Atomic number

= 173 -70

=103

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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.

makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

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What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

% => grams => moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H => 7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

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