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Artist 52 [7]
3 years ago
6

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>
  1. Learn more about neutrons brainly.com/question/1823582
  2. Learn more about nucleus brainly.com/question/2345919
  3. 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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PLEASE HELP!!
wariber [46]

Answer:

answer is option 2 hope it helps

8 0
3 years ago
Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g. The answer is assumed to be in mL.
Darya [45]

Answer:

0.0685 mL

Explanation:

To find the volume of the sample, divide the mass by the density.

(1.00 g)/(14.6 g/mL) = 0.0685 mL

3 0
3 years ago
Can you please help me?
lorasvet [3.4K]

Answer:

60 moles of NaF

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3

From the balanced equation above,

3 moles of NaF reacted to produce 1 mole of AlF3.

Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e

Xmol of NaF = 3 x 20

Xmol of NaF = 60 moles

Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.

4 0
3 years ago
How many grams of hydrogen are produced if 30.0 g of zinc reacts?
alekssr [168]
<span>0.925 grams if using hydrochloric acid in the reaction. 0.462 grams if using sulfuric acid in the reaction. 0.000 grams if using nitric acid in the reaction. Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is: Zn + 2 HCl ==> ZnCl2 + H2 So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen. Atomic weight zinc = 65.38 Atomic weight hydrogen = 1.00794 Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So 0.458855919 * 2 * 1.00794 = 0.92499847 grams. Rounding to 3 significant figures gives 0.925 grams. To show the assumption of the acid used, the balanced equation for sulfuric acid would be Zn2 + H2SO4 ==> Zn(SO4)2 + H2 Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid). If nitric acid were used, the reaction is 4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O Which means that NO hydrogen gas is generated. The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
3 0
3 years ago
How much volume (in cm^3)is gained by a person who gains 11.8 lb of our fat? Human fat has a density of 0.918 g/cm^3
Zinaida [17]

The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert pounds to grams </h3>

1 lb = 453.592 g

Therefore,

11.8 lb = 11.8 × 453.592

11.8 lb = 5352.3856 g

<h3>How to determine the volume </h3>
  • Mass = 5352.3856 g
  • Density = 0.918 g/cm³
  • Volume =?

Volume =  mass / density

Volume =  5352.3856 / 0.918

Volume = 5830.49 cm³

Learn more about density:

brainly.com/question/952755

#SPJ1

3 0
1 year ago
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