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Harman [31]
3 years ago
9

PLEASE HELP define selective breeding

Chemistry
1 answer:
Margarita [4]3 years ago
5 0
Hello, the answer to your question is simple

 Selective breeding (if we're talking about dogs in this example) always means the breeder is particular on not allowing: certain genentic mutations in the D.N.A, what color that animal is, if the animal is in good health and any genetic disorders.

   it is much more then choosing a partner

                Signed by, Virtouso Sargedog
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When a boron atom loses its valence electrons, what is the formula for the resulting ion?
kirill [66]

Answer:

option b. B3+

Explanation:

Boron takes the 5th position on the periodic table, therefore it has 5 electrons....2 on the inside and 3 on the outside. when it lost it 3 external electrons, it become positively charged with the amount of electron it loses.

8 0
2 years ago
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What is the size of Neptune and the state of matter?
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3 years ago
Please help! If a forensic chemist discovers evidence through examination which would prove a person guilty of a crime, why then
Bogdan [553]

Explanation: You cannot be convicted of a crime without evidence. You cannot be convicted of a state crime. You cannot be convicted of a federal crime. If there is no evidence against you, under the law, it simply is not possible for the prosecutor's office to obtain a conviction at trial. So, when you do have evidence it is really impossible to not get convicted unless the judge truly believes your side of the story and declares you not guilty.

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7 0
3 years ago
I bet no one can solve this
Nikitich [7]
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3 0
3 years ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
3 years ago
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