Why electrons present around the nucleus of an atom do not fall into the nucleus

Protons and neutrons are made up of smaller subatomic particles. When protons or neutrons get close enough to each other, they exchange particles (mesons), binding them together. Once they are bound, it takes considerable energy to break them apart.

7 0

3 years ago

A red-colored female New Hampshire Red chicken mated with a red-colored male New Hampshire Red chicken and produced golden-color

kifflom [539]

Answer:

See the answer below

Explanation:

The statement that could best explain how the offspring could inherit a trait not shown in the phenotypes of either parent would be that <u>both parents are heterozygous for the red color. </u>

Heterozygous parents can produce offspring with a phenotype that is distinctly different from that of either parent if the recessive alleles come together during random fusion of gametes.

Let us assume that the red color (R) in the New Hampshire chicken is the dominant one and the recessive one is the golden color (r). The genotype of the heterozygous parents would be Rr.

Rr x Rr

RR Rr Rr rr

<em>There is a probability (1/4) of producing a golden color chicken (rr), a trait that is not displayed in either of the parents.</em>

4 0

3 years ago

121 grams of iron (II) acetate reacts with 360 grams of lead (II) oxalate. Predict the products, write the complete reaction equ

egoroff_w [7]

Answer:

The mass of the precipitate is 100.2 grams

Explanation:

Step 1: Data given

Mass of iron (II) acetate = 121.0 grams

Mass of lead (II) oxalate = 360.0 grams

Molar mass Fe(C2H3O2)2 = 173.93 g/mol

Molar mass PbC2O4 = 295.22 g/mol

Step 2: The balanced equation

Fe(C2H3O2)2 + PbC2O4 → FeC2O4 + Pb(C2H3O2)2

Step 3: Calculate moles iron (II) acetate

Moles = mass / molar mass

Moles = 121.0 grams / 173.93 g/mol

Moles = 0.696 moles

Step 4: Calculate moles lead (II) oxalate

Moles = 360.0 grams / 295.22 g/mol

Moles = 1.219 moles

Step 5: Calculate limiting reactant

For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate

iron (II) acetate is the limiting reactant. It will completely be consumed ( 0.696 moles). Lead (II) oxalate is in excess. There will react 0.696 moles. There will remain 1.219 - 0.696 = 0.523 moles

Step 6: Calculate moles of iron(II) oxalate

For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron(II) oxalate and 1 mol Lead(II) acetate

For 0.696 moles iron(II) acetate we'll have 0.696 moles iron(II) oxalate

Step 7: Calculate mass iron(II) oxalate

Mass = moles * molar mass

Mass = 0.696 moles * 143.91 g/mol

Mass = 100.2 grams

The mass of the precipitate is 100.2 grams

7 0

3 years ago

Why electrons present around the nucleus of an atom do not fall into the nucleus​

Why electrons present around the nucleus of an atom do not fall into the nucleus